Evaluating an improper integral $int_{0}^{infty}frac{x^2}{(x^4+1)^2}dx$
I tried to solve the integral: $$int_{0}^{infty}frac{x^2}{(x^4+1)^2}dx$$
using $ x = sqrt{tan(u)}$ and $dx = frac{ sec^2(u)}{2sqrt{tan(u)}} du,$
but I ended up with an even worse looking integral $$ int_{0}^{frac{pi}{2}}frac{sqrt{tan(u)}}{sec^2(u)}du.$$
Wolfram gave an answer of $ dfrac{pi}{8sqrt{2}},$ but how would one get to that answer?
calculus definite-integrals improper-integrals
add a comment |
I tried to solve the integral: $$int_{0}^{infty}frac{x^2}{(x^4+1)^2}dx$$
using $ x = sqrt{tan(u)}$ and $dx = frac{ sec^2(u)}{2sqrt{tan(u)}} du,$
but I ended up with an even worse looking integral $$ int_{0}^{frac{pi}{2}}frac{sqrt{tan(u)}}{sec^2(u)}du.$$
Wolfram gave an answer of $ dfrac{pi}{8sqrt{2}},$ but how would one get to that answer?
calculus definite-integrals improper-integrals
Are you familiar with either the Beta function or the Residue Theorem?
– Zachary
Jan 4 at 1:29
@Zachary Unfortunately no I'm in bc calc and we haven't learned any of the fancier integration techniques
– Jessca
Jan 4 at 1:33
Jess you should be able to use partial fractions, using $$ x^4+1 = (x^2 + x sqrt 2 + 1) (x^2 - x sqrt 2 + 1)$$
– Will Jagy
Jan 4 at 1:41
add a comment |
I tried to solve the integral: $$int_{0}^{infty}frac{x^2}{(x^4+1)^2}dx$$
using $ x = sqrt{tan(u)}$ and $dx = frac{ sec^2(u)}{2sqrt{tan(u)}} du,$
but I ended up with an even worse looking integral $$ int_{0}^{frac{pi}{2}}frac{sqrt{tan(u)}}{sec^2(u)}du.$$
Wolfram gave an answer of $ dfrac{pi}{8sqrt{2}},$ but how would one get to that answer?
calculus definite-integrals improper-integrals
I tried to solve the integral: $$int_{0}^{infty}frac{x^2}{(x^4+1)^2}dx$$
using $ x = sqrt{tan(u)}$ and $dx = frac{ sec^2(u)}{2sqrt{tan(u)}} du,$
but I ended up with an even worse looking integral $$ int_{0}^{frac{pi}{2}}frac{sqrt{tan(u)}}{sec^2(u)}du.$$
Wolfram gave an answer of $ dfrac{pi}{8sqrt{2}},$ but how would one get to that answer?
calculus definite-integrals improper-integrals
calculus definite-integrals improper-integrals
edited 2 days ago
amWhy
192k28225439
192k28225439
asked Jan 4 at 1:22
JesscaJessca
455
455
Are you familiar with either the Beta function or the Residue Theorem?
– Zachary
Jan 4 at 1:29
@Zachary Unfortunately no I'm in bc calc and we haven't learned any of the fancier integration techniques
– Jessca
Jan 4 at 1:33
Jess you should be able to use partial fractions, using $$ x^4+1 = (x^2 + x sqrt 2 + 1) (x^2 - x sqrt 2 + 1)$$
– Will Jagy
Jan 4 at 1:41
add a comment |
Are you familiar with either the Beta function or the Residue Theorem?
– Zachary
Jan 4 at 1:29
@Zachary Unfortunately no I'm in bc calc and we haven't learned any of the fancier integration techniques
– Jessca
Jan 4 at 1:33
Jess you should be able to use partial fractions, using $$ x^4+1 = (x^2 + x sqrt 2 + 1) (x^2 - x sqrt 2 + 1)$$
– Will Jagy
Jan 4 at 1:41
Are you familiar with either the Beta function or the Residue Theorem?
– Zachary
Jan 4 at 1:29
Are you familiar with either the Beta function or the Residue Theorem?
– Zachary
Jan 4 at 1:29
@Zachary Unfortunately no I'm in bc calc and we haven't learned any of the fancier integration techniques
– Jessca
Jan 4 at 1:33
@Zachary Unfortunately no I'm in bc calc and we haven't learned any of the fancier integration techniques
– Jessca
Jan 4 at 1:33
Jess you should be able to use partial fractions, using $$ x^4+1 = (x^2 + x sqrt 2 + 1) (x^2 - x sqrt 2 + 1)$$
– Will Jagy
Jan 4 at 1:41
Jess you should be able to use partial fractions, using $$ x^4+1 = (x^2 + x sqrt 2 + 1) (x^2 - x sqrt 2 + 1)$$
– Will Jagy
Jan 4 at 1:41
add a comment |
5 Answers
5
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votes
Let us start with a step of integration by parts:
$$ int_{0}^{+infty}frac{1}{4x}cdotfrac{4x^3}{(x^4+1)^2},dx =int_{0}^{+infty}frac{1}{4x^2}left(1-frac{1}{1+x^4}right),dx=frac{1}{4}int_{0}^{+infty}frac{dx}{x^2+frac{1}{x^2}}$$
and finish with Glasser's master theorem:
$$ frac{1}{8}int_{-infty}^{+infty}frac{dx}{left(x-frac{1}{x}right)^2+2}stackrel{text{GMT}}{=}frac{1}{8}int_{-infty}^{+infty}frac{dx}{x^2+2} = frac{pi}{8sqrt{2}}.$$
1
I was actually just thinking about if it was possible to use GMT to evaluate that integral!
– Zachary
Jan 4 at 1:57
1
Quite impressive! Since the OP mentioned in the comments that hasn't learnt any fancier tricks I just want to mention that on the integral obtained after integrating by parts, we can do a $x=frac{1}{t}$ thus: $$J=int_0^infty frac{x^2} {1+x^4} dx=int_0^infty frac{dt} {1 +t^4}dt$$And if we add those two we get:$$2J=int_0^infty frac{x^2 +1}{x^4 +1}dx=int_0^infty frac{frac{1} {x^2} +1}{x^2 +frac{1} {x^2} }dx$$ And now completing the square in the denominator and letting $x-frac{1} {x} =t$ yields the same thing as by GMT.
– Zacky
Jan 4 at 2:03
I'm a little confused on how $int_{0}^{infty}frac{1}{4x}cdotfrac{4x^3}{(x^4+1)^2},dx =int_{0}^{infty}frac{1}{4x^2}left(1-frac{1}{1+x^4}right),dx$ after integrating by parts becuase I got $frac{1}{4x}(frac{-1}{x^4+1})vert_{0}^{infty} - int_{0}^{infty}frac{1}{4x^2(x^4+1)}$ how are these two equivalent?
– Jessca
Jan 4 at 2:26
2
@Jessca: it is better to take $1-frac{1}{1+x^4}$ as antiderivative of $frac{4x^3}{1+x^4}$, otherwise you end up with the difference of two divergent objects, not very useful.
– Jack D'Aurizio
Jan 4 at 2:40
add a comment |
Write $$frac{x^2}{(1+x^4)^2} = frac{4x^3}{(1+x^4)^2} cdot frac{1}{4x}.$$ Then integration by parts with the choice $$u = frac{1}{4x}, quad du = -frac{1}{4x^2} , dx, \ dv = frac{4x^3}{(1+x^4)^2} , dx, quad v = -frac{1}{1+x^4},$$ yields $$I_1(x) = int frac{x^2}{(1+x^4)^2} , dx = -frac{1}{4x(1+x^4)} - int frac{1}{4x^2(1+x^4)} , dx.$$ Now write $$frac{1}{x^2(1+x^4)} = frac{1}{x^2} - frac{x^2}{1+x^4},$$ thus $$I_1(x) = -frac{1}{4x(1+x^4)} + frac{1}{4x} + frac{1}{4} int frac{x^2}{1+x^4} , dx = frac{x^3}{4(1+x^4)} + frac{1}{4} I_2(x),$$ where we now seek to evaluate $I_2(x)$. This is accomplished in a number of ways; one is to perform the factorization $$1+x^4 = (1 + sqrt{2} x + x^2)(1 - sqrt{2} x + x^2),$$ and compute the partial fraction decomposition $$frac{x^2}{1+x^4} = frac{1}{2sqrt{2}} left( frac{x}{1 - sqrt{2} x + x^2} - frac{x}{1 + sqrt{2} x + x^2} right).$$ Then each term is integrable in closed form. This is not the easiest or most elegant approach, but it is perhaps the most elementary, requiring no knowledge beyond AP Calculus BC.
add a comment |
Consider the integral
$$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm{d}x$$
Applying the substitution $t=sin(x)^2$, we see that
$$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm{d}t$$
$$I(a,b)=frac12int_0^1t^{frac{a+1}2-1}(1-t)^{frac{b+1}2-1}mathrm{d}t$$
Recall the definition of the Beta function
$$mathrm B(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm{d}t=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
Where $Gamma(s)$ is the Gamma Function.
Hence we see that
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
From $$frac{tan(x)^{1/2}}{sec(x)^2}=sin(x)^{1/2}cos(x)^{3/2}$$
We see that your integral is
$$I(1/2,3/2)=frac{Gamma(3/4)Gamma(5/4)}{2Gamma(2)}$$
$$I(1/2,3/2)=frac{Gamma(1/4)Gamma(3/4)}{8}$$
And from
$$Gamma(s)Gamma(1-s)=fracpi{sinpi s},qquad snotinBbb Z$$
We have
$$I(1/2,3/2)=frac{pi}{8sqrt2}$$
1
seems fine, but you will need some tool to compute the last expression. Note that $Gamma(3/4)Gamma(5/4)=-frac14Gamma(-1/4)Gamma(5/4)$, then you can use the reflection formula
– Masacroso
Jan 4 at 1:54
I would maybe at least give the definition of the Gamma function the OP may not be familiar with it
– Zachary
Jan 4 at 1:55
1
you last expression seems wrong, note that $Gamma(2)=1$ and so $$frac{Gamma(3/4)Gamma(5/4)}{2Gamma(2)}=frac{Gamma(3/4)Gamma(1/4)}{8}$$ where we used the functional equation $frac14Gamma(1/4)=Gamma(1/4+1)=Gamma(5/4)$
– Masacroso
Jan 4 at 2:09
add a comment |
We could do it with contour integration.
take the contour from 0 to R along the real axis.
$int_0^R frac {x^2}{(x^4+1)^2} dx$
The quater circle.
$int_0^{frac pi 2} frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) dt$
$lim_limits{Rto infty} frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) = 0$
And down the imaginary axis.
$int_R^0 frac {(e^{frac {pi}{2} i} x)^2}{((e^{frac {pi}{2} i} x)^4+1)^2} (e^{frac {pi}{2} i}) dx\
int_R^0 frac {-x^2}{x^4+1)^2} (i) dx\
$
$(1+i)int_0^infty frac {x^2}{(x^4+1)^2} dx = (2pi i) text{ Res}_{left(x=e^{fracpi4i}right)}frac {x^2}{(x^4+1)^2}$
The pole is of order 2.
$frac {d}{dx}frac {x^2}{(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i})^2} = frac {2x(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i}) - 2x^2(3x^2 + 2xe^{frac pi4 i} + e^{frac {2pi}{4} i})}{(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i})^3}$
Evaluated at $e^{frac {pi}{4} i}$
$frac {4}{(4e^{frac {3pi}4 i})^3} = frac {1}{16e^{frac {pi}4 i}}$
$int_0^infty frac {x^2}{(x^4+1)^2} dx = frac {2pi i}{16sqrt 2 i} = frac {pi}{8sqrt 2}$
add a comment |
Using the method I employed here: we observe that
begin{equation}
int_{0}^{infty} frac{x^2}{left(x^4 + 1right)^2}:dx = frac{1}{4} cdot 1^{frac{2 + 1}{4} - 2} cdot Bleft(2 - frac{2 + 1}{4}, frac{2 + 1}{4} right) = frac{1}{4}Bleft(frac{5}{4}, frac{3}{4}right)
end{equation}
Using the relationship between the Beta and Gamma function:
begin{align}
int_{0}^{infty} frac{x^2}{left(x^4 + 1right)^2}:dx &= frac{1}{4}Bleft(frac{5}{4}, frac{3}{4}right) = frac{1}{4}frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{Gammaleft(frac{5}{4} + frac{3}{4}right)} \
&= frac{1}{4}frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{Gammaleft(2right)} =frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{8}
end{align}
add a comment |
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5 Answers
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5 Answers
5
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oldest
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Let us start with a step of integration by parts:
$$ int_{0}^{+infty}frac{1}{4x}cdotfrac{4x^3}{(x^4+1)^2},dx =int_{0}^{+infty}frac{1}{4x^2}left(1-frac{1}{1+x^4}right),dx=frac{1}{4}int_{0}^{+infty}frac{dx}{x^2+frac{1}{x^2}}$$
and finish with Glasser's master theorem:
$$ frac{1}{8}int_{-infty}^{+infty}frac{dx}{left(x-frac{1}{x}right)^2+2}stackrel{text{GMT}}{=}frac{1}{8}int_{-infty}^{+infty}frac{dx}{x^2+2} = frac{pi}{8sqrt{2}}.$$
1
I was actually just thinking about if it was possible to use GMT to evaluate that integral!
– Zachary
Jan 4 at 1:57
1
Quite impressive! Since the OP mentioned in the comments that hasn't learnt any fancier tricks I just want to mention that on the integral obtained after integrating by parts, we can do a $x=frac{1}{t}$ thus: $$J=int_0^infty frac{x^2} {1+x^4} dx=int_0^infty frac{dt} {1 +t^4}dt$$And if we add those two we get:$$2J=int_0^infty frac{x^2 +1}{x^4 +1}dx=int_0^infty frac{frac{1} {x^2} +1}{x^2 +frac{1} {x^2} }dx$$ And now completing the square in the denominator and letting $x-frac{1} {x} =t$ yields the same thing as by GMT.
– Zacky
Jan 4 at 2:03
I'm a little confused on how $int_{0}^{infty}frac{1}{4x}cdotfrac{4x^3}{(x^4+1)^2},dx =int_{0}^{infty}frac{1}{4x^2}left(1-frac{1}{1+x^4}right),dx$ after integrating by parts becuase I got $frac{1}{4x}(frac{-1}{x^4+1})vert_{0}^{infty} - int_{0}^{infty}frac{1}{4x^2(x^4+1)}$ how are these two equivalent?
– Jessca
Jan 4 at 2:26
2
@Jessca: it is better to take $1-frac{1}{1+x^4}$ as antiderivative of $frac{4x^3}{1+x^4}$, otherwise you end up with the difference of two divergent objects, not very useful.
– Jack D'Aurizio
Jan 4 at 2:40
add a comment |
Let us start with a step of integration by parts:
$$ int_{0}^{+infty}frac{1}{4x}cdotfrac{4x^3}{(x^4+1)^2},dx =int_{0}^{+infty}frac{1}{4x^2}left(1-frac{1}{1+x^4}right),dx=frac{1}{4}int_{0}^{+infty}frac{dx}{x^2+frac{1}{x^2}}$$
and finish with Glasser's master theorem:
$$ frac{1}{8}int_{-infty}^{+infty}frac{dx}{left(x-frac{1}{x}right)^2+2}stackrel{text{GMT}}{=}frac{1}{8}int_{-infty}^{+infty}frac{dx}{x^2+2} = frac{pi}{8sqrt{2}}.$$
1
I was actually just thinking about if it was possible to use GMT to evaluate that integral!
– Zachary
Jan 4 at 1:57
1
Quite impressive! Since the OP mentioned in the comments that hasn't learnt any fancier tricks I just want to mention that on the integral obtained after integrating by parts, we can do a $x=frac{1}{t}$ thus: $$J=int_0^infty frac{x^2} {1+x^4} dx=int_0^infty frac{dt} {1 +t^4}dt$$And if we add those two we get:$$2J=int_0^infty frac{x^2 +1}{x^4 +1}dx=int_0^infty frac{frac{1} {x^2} +1}{x^2 +frac{1} {x^2} }dx$$ And now completing the square in the denominator and letting $x-frac{1} {x} =t$ yields the same thing as by GMT.
– Zacky
Jan 4 at 2:03
I'm a little confused on how $int_{0}^{infty}frac{1}{4x}cdotfrac{4x^3}{(x^4+1)^2},dx =int_{0}^{infty}frac{1}{4x^2}left(1-frac{1}{1+x^4}right),dx$ after integrating by parts becuase I got $frac{1}{4x}(frac{-1}{x^4+1})vert_{0}^{infty} - int_{0}^{infty}frac{1}{4x^2(x^4+1)}$ how are these two equivalent?
– Jessca
Jan 4 at 2:26
2
@Jessca: it is better to take $1-frac{1}{1+x^4}$ as antiderivative of $frac{4x^3}{1+x^4}$, otherwise you end up with the difference of two divergent objects, not very useful.
– Jack D'Aurizio
Jan 4 at 2:40
add a comment |
Let us start with a step of integration by parts:
$$ int_{0}^{+infty}frac{1}{4x}cdotfrac{4x^3}{(x^4+1)^2},dx =int_{0}^{+infty}frac{1}{4x^2}left(1-frac{1}{1+x^4}right),dx=frac{1}{4}int_{0}^{+infty}frac{dx}{x^2+frac{1}{x^2}}$$
and finish with Glasser's master theorem:
$$ frac{1}{8}int_{-infty}^{+infty}frac{dx}{left(x-frac{1}{x}right)^2+2}stackrel{text{GMT}}{=}frac{1}{8}int_{-infty}^{+infty}frac{dx}{x^2+2} = frac{pi}{8sqrt{2}}.$$
Let us start with a step of integration by parts:
$$ int_{0}^{+infty}frac{1}{4x}cdotfrac{4x^3}{(x^4+1)^2},dx =int_{0}^{+infty}frac{1}{4x^2}left(1-frac{1}{1+x^4}right),dx=frac{1}{4}int_{0}^{+infty}frac{dx}{x^2+frac{1}{x^2}}$$
and finish with Glasser's master theorem:
$$ frac{1}{8}int_{-infty}^{+infty}frac{dx}{left(x-frac{1}{x}right)^2+2}stackrel{text{GMT}}{=}frac{1}{8}int_{-infty}^{+infty}frac{dx}{x^2+2} = frac{pi}{8sqrt{2}}.$$
edited Jan 4 at 1:38
answered Jan 4 at 1:26
Jack D'AurizioJack D'Aurizio
287k33280658
287k33280658
1
I was actually just thinking about if it was possible to use GMT to evaluate that integral!
– Zachary
Jan 4 at 1:57
1
Quite impressive! Since the OP mentioned in the comments that hasn't learnt any fancier tricks I just want to mention that on the integral obtained after integrating by parts, we can do a $x=frac{1}{t}$ thus: $$J=int_0^infty frac{x^2} {1+x^4} dx=int_0^infty frac{dt} {1 +t^4}dt$$And if we add those two we get:$$2J=int_0^infty frac{x^2 +1}{x^4 +1}dx=int_0^infty frac{frac{1} {x^2} +1}{x^2 +frac{1} {x^2} }dx$$ And now completing the square in the denominator and letting $x-frac{1} {x} =t$ yields the same thing as by GMT.
– Zacky
Jan 4 at 2:03
I'm a little confused on how $int_{0}^{infty}frac{1}{4x}cdotfrac{4x^3}{(x^4+1)^2},dx =int_{0}^{infty}frac{1}{4x^2}left(1-frac{1}{1+x^4}right),dx$ after integrating by parts becuase I got $frac{1}{4x}(frac{-1}{x^4+1})vert_{0}^{infty} - int_{0}^{infty}frac{1}{4x^2(x^4+1)}$ how are these two equivalent?
– Jessca
Jan 4 at 2:26
2
@Jessca: it is better to take $1-frac{1}{1+x^4}$ as antiderivative of $frac{4x^3}{1+x^4}$, otherwise you end up with the difference of two divergent objects, not very useful.
– Jack D'Aurizio
Jan 4 at 2:40
add a comment |
1
I was actually just thinking about if it was possible to use GMT to evaluate that integral!
– Zachary
Jan 4 at 1:57
1
Quite impressive! Since the OP mentioned in the comments that hasn't learnt any fancier tricks I just want to mention that on the integral obtained after integrating by parts, we can do a $x=frac{1}{t}$ thus: $$J=int_0^infty frac{x^2} {1+x^4} dx=int_0^infty frac{dt} {1 +t^4}dt$$And if we add those two we get:$$2J=int_0^infty frac{x^2 +1}{x^4 +1}dx=int_0^infty frac{frac{1} {x^2} +1}{x^2 +frac{1} {x^2} }dx$$ And now completing the square in the denominator and letting $x-frac{1} {x} =t$ yields the same thing as by GMT.
– Zacky
Jan 4 at 2:03
I'm a little confused on how $int_{0}^{infty}frac{1}{4x}cdotfrac{4x^3}{(x^4+1)^2},dx =int_{0}^{infty}frac{1}{4x^2}left(1-frac{1}{1+x^4}right),dx$ after integrating by parts becuase I got $frac{1}{4x}(frac{-1}{x^4+1})vert_{0}^{infty} - int_{0}^{infty}frac{1}{4x^2(x^4+1)}$ how are these two equivalent?
– Jessca
Jan 4 at 2:26
2
@Jessca: it is better to take $1-frac{1}{1+x^4}$ as antiderivative of $frac{4x^3}{1+x^4}$, otherwise you end up with the difference of two divergent objects, not very useful.
– Jack D'Aurizio
Jan 4 at 2:40
1
1
I was actually just thinking about if it was possible to use GMT to evaluate that integral!
– Zachary
Jan 4 at 1:57
I was actually just thinking about if it was possible to use GMT to evaluate that integral!
– Zachary
Jan 4 at 1:57
1
1
Quite impressive! Since the OP mentioned in the comments that hasn't learnt any fancier tricks I just want to mention that on the integral obtained after integrating by parts, we can do a $x=frac{1}{t}$ thus: $$J=int_0^infty frac{x^2} {1+x^4} dx=int_0^infty frac{dt} {1 +t^4}dt$$And if we add those two we get:$$2J=int_0^infty frac{x^2 +1}{x^4 +1}dx=int_0^infty frac{frac{1} {x^2} +1}{x^2 +frac{1} {x^2} }dx$$ And now completing the square in the denominator and letting $x-frac{1} {x} =t$ yields the same thing as by GMT.
– Zacky
Jan 4 at 2:03
Quite impressive! Since the OP mentioned in the comments that hasn't learnt any fancier tricks I just want to mention that on the integral obtained after integrating by parts, we can do a $x=frac{1}{t}$ thus: $$J=int_0^infty frac{x^2} {1+x^4} dx=int_0^infty frac{dt} {1 +t^4}dt$$And if we add those two we get:$$2J=int_0^infty frac{x^2 +1}{x^4 +1}dx=int_0^infty frac{frac{1} {x^2} +1}{x^2 +frac{1} {x^2} }dx$$ And now completing the square in the denominator and letting $x-frac{1} {x} =t$ yields the same thing as by GMT.
– Zacky
Jan 4 at 2:03
I'm a little confused on how $int_{0}^{infty}frac{1}{4x}cdotfrac{4x^3}{(x^4+1)^2},dx =int_{0}^{infty}frac{1}{4x^2}left(1-frac{1}{1+x^4}right),dx$ after integrating by parts becuase I got $frac{1}{4x}(frac{-1}{x^4+1})vert_{0}^{infty} - int_{0}^{infty}frac{1}{4x^2(x^4+1)}$ how are these two equivalent?
– Jessca
Jan 4 at 2:26
I'm a little confused on how $int_{0}^{infty}frac{1}{4x}cdotfrac{4x^3}{(x^4+1)^2},dx =int_{0}^{infty}frac{1}{4x^2}left(1-frac{1}{1+x^4}right),dx$ after integrating by parts becuase I got $frac{1}{4x}(frac{-1}{x^4+1})vert_{0}^{infty} - int_{0}^{infty}frac{1}{4x^2(x^4+1)}$ how are these two equivalent?
– Jessca
Jan 4 at 2:26
2
2
@Jessca: it is better to take $1-frac{1}{1+x^4}$ as antiderivative of $frac{4x^3}{1+x^4}$, otherwise you end up with the difference of two divergent objects, not very useful.
– Jack D'Aurizio
Jan 4 at 2:40
@Jessca: it is better to take $1-frac{1}{1+x^4}$ as antiderivative of $frac{4x^3}{1+x^4}$, otherwise you end up with the difference of two divergent objects, not very useful.
– Jack D'Aurizio
Jan 4 at 2:40
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Write $$frac{x^2}{(1+x^4)^2} = frac{4x^3}{(1+x^4)^2} cdot frac{1}{4x}.$$ Then integration by parts with the choice $$u = frac{1}{4x}, quad du = -frac{1}{4x^2} , dx, \ dv = frac{4x^3}{(1+x^4)^2} , dx, quad v = -frac{1}{1+x^4},$$ yields $$I_1(x) = int frac{x^2}{(1+x^4)^2} , dx = -frac{1}{4x(1+x^4)} - int frac{1}{4x^2(1+x^4)} , dx.$$ Now write $$frac{1}{x^2(1+x^4)} = frac{1}{x^2} - frac{x^2}{1+x^4},$$ thus $$I_1(x) = -frac{1}{4x(1+x^4)} + frac{1}{4x} + frac{1}{4} int frac{x^2}{1+x^4} , dx = frac{x^3}{4(1+x^4)} + frac{1}{4} I_2(x),$$ where we now seek to evaluate $I_2(x)$. This is accomplished in a number of ways; one is to perform the factorization $$1+x^4 = (1 + sqrt{2} x + x^2)(1 - sqrt{2} x + x^2),$$ and compute the partial fraction decomposition $$frac{x^2}{1+x^4} = frac{1}{2sqrt{2}} left( frac{x}{1 - sqrt{2} x + x^2} - frac{x}{1 + sqrt{2} x + x^2} right).$$ Then each term is integrable in closed form. This is not the easiest or most elegant approach, but it is perhaps the most elementary, requiring no knowledge beyond AP Calculus BC.
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Write $$frac{x^2}{(1+x^4)^2} = frac{4x^3}{(1+x^4)^2} cdot frac{1}{4x}.$$ Then integration by parts with the choice $$u = frac{1}{4x}, quad du = -frac{1}{4x^2} , dx, \ dv = frac{4x^3}{(1+x^4)^2} , dx, quad v = -frac{1}{1+x^4},$$ yields $$I_1(x) = int frac{x^2}{(1+x^4)^2} , dx = -frac{1}{4x(1+x^4)} - int frac{1}{4x^2(1+x^4)} , dx.$$ Now write $$frac{1}{x^2(1+x^4)} = frac{1}{x^2} - frac{x^2}{1+x^4},$$ thus $$I_1(x) = -frac{1}{4x(1+x^4)} + frac{1}{4x} + frac{1}{4} int frac{x^2}{1+x^4} , dx = frac{x^3}{4(1+x^4)} + frac{1}{4} I_2(x),$$ where we now seek to evaluate $I_2(x)$. This is accomplished in a number of ways; one is to perform the factorization $$1+x^4 = (1 + sqrt{2} x + x^2)(1 - sqrt{2} x + x^2),$$ and compute the partial fraction decomposition $$frac{x^2}{1+x^4} = frac{1}{2sqrt{2}} left( frac{x}{1 - sqrt{2} x + x^2} - frac{x}{1 + sqrt{2} x + x^2} right).$$ Then each term is integrable in closed form. This is not the easiest or most elegant approach, but it is perhaps the most elementary, requiring no knowledge beyond AP Calculus BC.
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Write $$frac{x^2}{(1+x^4)^2} = frac{4x^3}{(1+x^4)^2} cdot frac{1}{4x}.$$ Then integration by parts with the choice $$u = frac{1}{4x}, quad du = -frac{1}{4x^2} , dx, \ dv = frac{4x^3}{(1+x^4)^2} , dx, quad v = -frac{1}{1+x^4},$$ yields $$I_1(x) = int frac{x^2}{(1+x^4)^2} , dx = -frac{1}{4x(1+x^4)} - int frac{1}{4x^2(1+x^4)} , dx.$$ Now write $$frac{1}{x^2(1+x^4)} = frac{1}{x^2} - frac{x^2}{1+x^4},$$ thus $$I_1(x) = -frac{1}{4x(1+x^4)} + frac{1}{4x} + frac{1}{4} int frac{x^2}{1+x^4} , dx = frac{x^3}{4(1+x^4)} + frac{1}{4} I_2(x),$$ where we now seek to evaluate $I_2(x)$. This is accomplished in a number of ways; one is to perform the factorization $$1+x^4 = (1 + sqrt{2} x + x^2)(1 - sqrt{2} x + x^2),$$ and compute the partial fraction decomposition $$frac{x^2}{1+x^4} = frac{1}{2sqrt{2}} left( frac{x}{1 - sqrt{2} x + x^2} - frac{x}{1 + sqrt{2} x + x^2} right).$$ Then each term is integrable in closed form. This is not the easiest or most elegant approach, but it is perhaps the most elementary, requiring no knowledge beyond AP Calculus BC.
Write $$frac{x^2}{(1+x^4)^2} = frac{4x^3}{(1+x^4)^2} cdot frac{1}{4x}.$$ Then integration by parts with the choice $$u = frac{1}{4x}, quad du = -frac{1}{4x^2} , dx, \ dv = frac{4x^3}{(1+x^4)^2} , dx, quad v = -frac{1}{1+x^4},$$ yields $$I_1(x) = int frac{x^2}{(1+x^4)^2} , dx = -frac{1}{4x(1+x^4)} - int frac{1}{4x^2(1+x^4)} , dx.$$ Now write $$frac{1}{x^2(1+x^4)} = frac{1}{x^2} - frac{x^2}{1+x^4},$$ thus $$I_1(x) = -frac{1}{4x(1+x^4)} + frac{1}{4x} + frac{1}{4} int frac{x^2}{1+x^4} , dx = frac{x^3}{4(1+x^4)} + frac{1}{4} I_2(x),$$ where we now seek to evaluate $I_2(x)$. This is accomplished in a number of ways; one is to perform the factorization $$1+x^4 = (1 + sqrt{2} x + x^2)(1 - sqrt{2} x + x^2),$$ and compute the partial fraction decomposition $$frac{x^2}{1+x^4} = frac{1}{2sqrt{2}} left( frac{x}{1 - sqrt{2} x + x^2} - frac{x}{1 + sqrt{2} x + x^2} right).$$ Then each term is integrable in closed form. This is not the easiest or most elegant approach, but it is perhaps the most elementary, requiring no knowledge beyond AP Calculus BC.
answered Jan 4 at 1:54
heropupheropup
62.6k66099
62.6k66099
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Consider the integral
$$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm{d}x$$
Applying the substitution $t=sin(x)^2$, we see that
$$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm{d}t$$
$$I(a,b)=frac12int_0^1t^{frac{a+1}2-1}(1-t)^{frac{b+1}2-1}mathrm{d}t$$
Recall the definition of the Beta function
$$mathrm B(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm{d}t=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
Where $Gamma(s)$ is the Gamma Function.
Hence we see that
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
From $$frac{tan(x)^{1/2}}{sec(x)^2}=sin(x)^{1/2}cos(x)^{3/2}$$
We see that your integral is
$$I(1/2,3/2)=frac{Gamma(3/4)Gamma(5/4)}{2Gamma(2)}$$
$$I(1/2,3/2)=frac{Gamma(1/4)Gamma(3/4)}{8}$$
And from
$$Gamma(s)Gamma(1-s)=fracpi{sinpi s},qquad snotinBbb Z$$
We have
$$I(1/2,3/2)=frac{pi}{8sqrt2}$$
1
seems fine, but you will need some tool to compute the last expression. Note that $Gamma(3/4)Gamma(5/4)=-frac14Gamma(-1/4)Gamma(5/4)$, then you can use the reflection formula
– Masacroso
Jan 4 at 1:54
I would maybe at least give the definition of the Gamma function the OP may not be familiar with it
– Zachary
Jan 4 at 1:55
1
you last expression seems wrong, note that $Gamma(2)=1$ and so $$frac{Gamma(3/4)Gamma(5/4)}{2Gamma(2)}=frac{Gamma(3/4)Gamma(1/4)}{8}$$ where we used the functional equation $frac14Gamma(1/4)=Gamma(1/4+1)=Gamma(5/4)$
– Masacroso
Jan 4 at 2:09
add a comment |
Consider the integral
$$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm{d}x$$
Applying the substitution $t=sin(x)^2$, we see that
$$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm{d}t$$
$$I(a,b)=frac12int_0^1t^{frac{a+1}2-1}(1-t)^{frac{b+1}2-1}mathrm{d}t$$
Recall the definition of the Beta function
$$mathrm B(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm{d}t=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
Where $Gamma(s)$ is the Gamma Function.
Hence we see that
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
From $$frac{tan(x)^{1/2}}{sec(x)^2}=sin(x)^{1/2}cos(x)^{3/2}$$
We see that your integral is
$$I(1/2,3/2)=frac{Gamma(3/4)Gamma(5/4)}{2Gamma(2)}$$
$$I(1/2,3/2)=frac{Gamma(1/4)Gamma(3/4)}{8}$$
And from
$$Gamma(s)Gamma(1-s)=fracpi{sinpi s},qquad snotinBbb Z$$
We have
$$I(1/2,3/2)=frac{pi}{8sqrt2}$$
1
seems fine, but you will need some tool to compute the last expression. Note that $Gamma(3/4)Gamma(5/4)=-frac14Gamma(-1/4)Gamma(5/4)$, then you can use the reflection formula
– Masacroso
Jan 4 at 1:54
I would maybe at least give the definition of the Gamma function the OP may not be familiar with it
– Zachary
Jan 4 at 1:55
1
you last expression seems wrong, note that $Gamma(2)=1$ and so $$frac{Gamma(3/4)Gamma(5/4)}{2Gamma(2)}=frac{Gamma(3/4)Gamma(1/4)}{8}$$ where we used the functional equation $frac14Gamma(1/4)=Gamma(1/4+1)=Gamma(5/4)$
– Masacroso
Jan 4 at 2:09
add a comment |
Consider the integral
$$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm{d}x$$
Applying the substitution $t=sin(x)^2$, we see that
$$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm{d}t$$
$$I(a,b)=frac12int_0^1t^{frac{a+1}2-1}(1-t)^{frac{b+1}2-1}mathrm{d}t$$
Recall the definition of the Beta function
$$mathrm B(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm{d}t=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
Where $Gamma(s)$ is the Gamma Function.
Hence we see that
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
From $$frac{tan(x)^{1/2}}{sec(x)^2}=sin(x)^{1/2}cos(x)^{3/2}$$
We see that your integral is
$$I(1/2,3/2)=frac{Gamma(3/4)Gamma(5/4)}{2Gamma(2)}$$
$$I(1/2,3/2)=frac{Gamma(1/4)Gamma(3/4)}{8}$$
And from
$$Gamma(s)Gamma(1-s)=fracpi{sinpi s},qquad snotinBbb Z$$
We have
$$I(1/2,3/2)=frac{pi}{8sqrt2}$$
Consider the integral
$$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm{d}x$$
Applying the substitution $t=sin(x)^2$, we see that
$$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm{d}t$$
$$I(a,b)=frac12int_0^1t^{frac{a+1}2-1}(1-t)^{frac{b+1}2-1}mathrm{d}t$$
Recall the definition of the Beta function
$$mathrm B(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm{d}t=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
Where $Gamma(s)$ is the Gamma Function.
Hence we see that
$$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
From $$frac{tan(x)^{1/2}}{sec(x)^2}=sin(x)^{1/2}cos(x)^{3/2}$$
We see that your integral is
$$I(1/2,3/2)=frac{Gamma(3/4)Gamma(5/4)}{2Gamma(2)}$$
$$I(1/2,3/2)=frac{Gamma(1/4)Gamma(3/4)}{8}$$
And from
$$Gamma(s)Gamma(1-s)=fracpi{sinpi s},qquad snotinBbb Z$$
We have
$$I(1/2,3/2)=frac{pi}{8sqrt2}$$
edited yesterday
answered Jan 4 at 1:40
clathratusclathratus
3,243331
3,243331
1
seems fine, but you will need some tool to compute the last expression. Note that $Gamma(3/4)Gamma(5/4)=-frac14Gamma(-1/4)Gamma(5/4)$, then you can use the reflection formula
– Masacroso
Jan 4 at 1:54
I would maybe at least give the definition of the Gamma function the OP may not be familiar with it
– Zachary
Jan 4 at 1:55
1
you last expression seems wrong, note that $Gamma(2)=1$ and so $$frac{Gamma(3/4)Gamma(5/4)}{2Gamma(2)}=frac{Gamma(3/4)Gamma(1/4)}{8}$$ where we used the functional equation $frac14Gamma(1/4)=Gamma(1/4+1)=Gamma(5/4)$
– Masacroso
Jan 4 at 2:09
add a comment |
1
seems fine, but you will need some tool to compute the last expression. Note that $Gamma(3/4)Gamma(5/4)=-frac14Gamma(-1/4)Gamma(5/4)$, then you can use the reflection formula
– Masacroso
Jan 4 at 1:54
I would maybe at least give the definition of the Gamma function the OP may not be familiar with it
– Zachary
Jan 4 at 1:55
1
you last expression seems wrong, note that $Gamma(2)=1$ and so $$frac{Gamma(3/4)Gamma(5/4)}{2Gamma(2)}=frac{Gamma(3/4)Gamma(1/4)}{8}$$ where we used the functional equation $frac14Gamma(1/4)=Gamma(1/4+1)=Gamma(5/4)$
– Masacroso
Jan 4 at 2:09
1
1
seems fine, but you will need some tool to compute the last expression. Note that $Gamma(3/4)Gamma(5/4)=-frac14Gamma(-1/4)Gamma(5/4)$, then you can use the reflection formula
– Masacroso
Jan 4 at 1:54
seems fine, but you will need some tool to compute the last expression. Note that $Gamma(3/4)Gamma(5/4)=-frac14Gamma(-1/4)Gamma(5/4)$, then you can use the reflection formula
– Masacroso
Jan 4 at 1:54
I would maybe at least give the definition of the Gamma function the OP may not be familiar with it
– Zachary
Jan 4 at 1:55
I would maybe at least give the definition of the Gamma function the OP may not be familiar with it
– Zachary
Jan 4 at 1:55
1
1
you last expression seems wrong, note that $Gamma(2)=1$ and so $$frac{Gamma(3/4)Gamma(5/4)}{2Gamma(2)}=frac{Gamma(3/4)Gamma(1/4)}{8}$$ where we used the functional equation $frac14Gamma(1/4)=Gamma(1/4+1)=Gamma(5/4)$
– Masacroso
Jan 4 at 2:09
you last expression seems wrong, note that $Gamma(2)=1$ and so $$frac{Gamma(3/4)Gamma(5/4)}{2Gamma(2)}=frac{Gamma(3/4)Gamma(1/4)}{8}$$ where we used the functional equation $frac14Gamma(1/4)=Gamma(1/4+1)=Gamma(5/4)$
– Masacroso
Jan 4 at 2:09
add a comment |
We could do it with contour integration.
take the contour from 0 to R along the real axis.
$int_0^R frac {x^2}{(x^4+1)^2} dx$
The quater circle.
$int_0^{frac pi 2} frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) dt$
$lim_limits{Rto infty} frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) = 0$
And down the imaginary axis.
$int_R^0 frac {(e^{frac {pi}{2} i} x)^2}{((e^{frac {pi}{2} i} x)^4+1)^2} (e^{frac {pi}{2} i}) dx\
int_R^0 frac {-x^2}{x^4+1)^2} (i) dx\
$
$(1+i)int_0^infty frac {x^2}{(x^4+1)^2} dx = (2pi i) text{ Res}_{left(x=e^{fracpi4i}right)}frac {x^2}{(x^4+1)^2}$
The pole is of order 2.
$frac {d}{dx}frac {x^2}{(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i})^2} = frac {2x(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i}) - 2x^2(3x^2 + 2xe^{frac pi4 i} + e^{frac {2pi}{4} i})}{(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i})^3}$
Evaluated at $e^{frac {pi}{4} i}$
$frac {4}{(4e^{frac {3pi}4 i})^3} = frac {1}{16e^{frac {pi}4 i}}$
$int_0^infty frac {x^2}{(x^4+1)^2} dx = frac {2pi i}{16sqrt 2 i} = frac {pi}{8sqrt 2}$
add a comment |
We could do it with contour integration.
take the contour from 0 to R along the real axis.
$int_0^R frac {x^2}{(x^4+1)^2} dx$
The quater circle.
$int_0^{frac pi 2} frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) dt$
$lim_limits{Rto infty} frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) = 0$
And down the imaginary axis.
$int_R^0 frac {(e^{frac {pi}{2} i} x)^2}{((e^{frac {pi}{2} i} x)^4+1)^2} (e^{frac {pi}{2} i}) dx\
int_R^0 frac {-x^2}{x^4+1)^2} (i) dx\
$
$(1+i)int_0^infty frac {x^2}{(x^4+1)^2} dx = (2pi i) text{ Res}_{left(x=e^{fracpi4i}right)}frac {x^2}{(x^4+1)^2}$
The pole is of order 2.
$frac {d}{dx}frac {x^2}{(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i})^2} = frac {2x(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i}) - 2x^2(3x^2 + 2xe^{frac pi4 i} + e^{frac {2pi}{4} i})}{(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i})^3}$
Evaluated at $e^{frac {pi}{4} i}$
$frac {4}{(4e^{frac {3pi}4 i})^3} = frac {1}{16e^{frac {pi}4 i}}$
$int_0^infty frac {x^2}{(x^4+1)^2} dx = frac {2pi i}{16sqrt 2 i} = frac {pi}{8sqrt 2}$
add a comment |
We could do it with contour integration.
take the contour from 0 to R along the real axis.
$int_0^R frac {x^2}{(x^4+1)^2} dx$
The quater circle.
$int_0^{frac pi 2} frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) dt$
$lim_limits{Rto infty} frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) = 0$
And down the imaginary axis.
$int_R^0 frac {(e^{frac {pi}{2} i} x)^2}{((e^{frac {pi}{2} i} x)^4+1)^2} (e^{frac {pi}{2} i}) dx\
int_R^0 frac {-x^2}{x^4+1)^2} (i) dx\
$
$(1+i)int_0^infty frac {x^2}{(x^4+1)^2} dx = (2pi i) text{ Res}_{left(x=e^{fracpi4i}right)}frac {x^2}{(x^4+1)^2}$
The pole is of order 2.
$frac {d}{dx}frac {x^2}{(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i})^2} = frac {2x(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i}) - 2x^2(3x^2 + 2xe^{frac pi4 i} + e^{frac {2pi}{4} i})}{(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i})^3}$
Evaluated at $e^{frac {pi}{4} i}$
$frac {4}{(4e^{frac {3pi}4 i})^3} = frac {1}{16e^{frac {pi}4 i}}$
$int_0^infty frac {x^2}{(x^4+1)^2} dx = frac {2pi i}{16sqrt 2 i} = frac {pi}{8sqrt 2}$
We could do it with contour integration.
take the contour from 0 to R along the real axis.
$int_0^R frac {x^2}{(x^4+1)^2} dx$
The quater circle.
$int_0^{frac pi 2} frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) dt$
$lim_limits{Rto infty} frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) = 0$
And down the imaginary axis.
$int_R^0 frac {(e^{frac {pi}{2} i} x)^2}{((e^{frac {pi}{2} i} x)^4+1)^2} (e^{frac {pi}{2} i}) dx\
int_R^0 frac {-x^2}{x^4+1)^2} (i) dx\
$
$(1+i)int_0^infty frac {x^2}{(x^4+1)^2} dx = (2pi i) text{ Res}_{left(x=e^{fracpi4i}right)}frac {x^2}{(x^4+1)^2}$
The pole is of order 2.
$frac {d}{dx}frac {x^2}{(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i})^2} = frac {2x(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i}) - 2x^2(3x^2 + 2xe^{frac pi4 i} + e^{frac {2pi}{4} i})}{(x^3 + x^2e^{frac pi4 i} + xe^{frac {2pi}{4} i}+e^{frac {3pi}{4} i})^3}$
Evaluated at $e^{frac {pi}{4} i}$
$frac {4}{(4e^{frac {3pi}4 i})^3} = frac {1}{16e^{frac {pi}4 i}}$
$int_0^infty frac {x^2}{(x^4+1)^2} dx = frac {2pi i}{16sqrt 2 i} = frac {pi}{8sqrt 2}$
answered Jan 4 at 2:29
Doug MDoug M
44.2k31854
44.2k31854
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Using the method I employed here: we observe that
begin{equation}
int_{0}^{infty} frac{x^2}{left(x^4 + 1right)^2}:dx = frac{1}{4} cdot 1^{frac{2 + 1}{4} - 2} cdot Bleft(2 - frac{2 + 1}{4}, frac{2 + 1}{4} right) = frac{1}{4}Bleft(frac{5}{4}, frac{3}{4}right)
end{equation}
Using the relationship between the Beta and Gamma function:
begin{align}
int_{0}^{infty} frac{x^2}{left(x^4 + 1right)^2}:dx &= frac{1}{4}Bleft(frac{5}{4}, frac{3}{4}right) = frac{1}{4}frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{Gammaleft(frac{5}{4} + frac{3}{4}right)} \
&= frac{1}{4}frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{Gammaleft(2right)} =frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{8}
end{align}
add a comment |
Using the method I employed here: we observe that
begin{equation}
int_{0}^{infty} frac{x^2}{left(x^4 + 1right)^2}:dx = frac{1}{4} cdot 1^{frac{2 + 1}{4} - 2} cdot Bleft(2 - frac{2 + 1}{4}, frac{2 + 1}{4} right) = frac{1}{4}Bleft(frac{5}{4}, frac{3}{4}right)
end{equation}
Using the relationship between the Beta and Gamma function:
begin{align}
int_{0}^{infty} frac{x^2}{left(x^4 + 1right)^2}:dx &= frac{1}{4}Bleft(frac{5}{4}, frac{3}{4}right) = frac{1}{4}frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{Gammaleft(frac{5}{4} + frac{3}{4}right)} \
&= frac{1}{4}frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{Gammaleft(2right)} =frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{8}
end{align}
add a comment |
Using the method I employed here: we observe that
begin{equation}
int_{0}^{infty} frac{x^2}{left(x^4 + 1right)^2}:dx = frac{1}{4} cdot 1^{frac{2 + 1}{4} - 2} cdot Bleft(2 - frac{2 + 1}{4}, frac{2 + 1}{4} right) = frac{1}{4}Bleft(frac{5}{4}, frac{3}{4}right)
end{equation}
Using the relationship between the Beta and Gamma function:
begin{align}
int_{0}^{infty} frac{x^2}{left(x^4 + 1right)^2}:dx &= frac{1}{4}Bleft(frac{5}{4}, frac{3}{4}right) = frac{1}{4}frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{Gammaleft(frac{5}{4} + frac{3}{4}right)} \
&= frac{1}{4}frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{Gammaleft(2right)} =frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{8}
end{align}
Using the method I employed here: we observe that
begin{equation}
int_{0}^{infty} frac{x^2}{left(x^4 + 1right)^2}:dx = frac{1}{4} cdot 1^{frac{2 + 1}{4} - 2} cdot Bleft(2 - frac{2 + 1}{4}, frac{2 + 1}{4} right) = frac{1}{4}Bleft(frac{5}{4}, frac{3}{4}right)
end{equation}
Using the relationship between the Beta and Gamma function:
begin{align}
int_{0}^{infty} frac{x^2}{left(x^4 + 1right)^2}:dx &= frac{1}{4}Bleft(frac{5}{4}, frac{3}{4}right) = frac{1}{4}frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{Gammaleft(frac{5}{4} + frac{3}{4}right)} \
&= frac{1}{4}frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{Gammaleft(2right)} =frac{Gammaleft(frac{5}{4}right)Gammaleft(frac{3}{4}right)}{8}
end{align}
answered Jan 4 at 8:00
DavidGDavidG
1,830620
1,830620
add a comment |
add a comment |
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Are you familiar with either the Beta function or the Residue Theorem?
– Zachary
Jan 4 at 1:29
@Zachary Unfortunately no I'm in bc calc and we haven't learned any of the fancier integration techniques
– Jessca
Jan 4 at 1:33
Jess you should be able to use partial fractions, using $$ x^4+1 = (x^2 + x sqrt 2 + 1) (x^2 - x sqrt 2 + 1)$$
– Will Jagy
Jan 4 at 1:41