Suppose $ alpha, beta>0 $. Compute: $ int_{0}^{infty}frac{cos (alpha x)-cos (beta x)}{x}dx $












8















Suppose $ alpha, beta>0 $. Compute:
$$ int_{0}^{infty}frac{cos (alpha x)-cos (beta x)}{x}dx $$




Here is what I do:
$$begin{align}
int_{0}^{infty}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}sin (yx)dy\
&=int_{alpha}^{beta}dyint_{0}^{infty}sin(yx)dx\
& \
& qquadtext{let $ yx=u $}\
& \
&=int_{alpha}^{beta}frac{1}{y}dyint_0^{infty}sin u du\
&=int_{alpha}^{beta}frac{1}{y}dyleft( -cos u|_{infty}+cos u|_0 right)\
&=logfrac{beta}{alpha}(-cos(infty)+1)\
&=logfrac{beta}{alpha}-cos(infty)logfrac{beta}{alpha}
end{align}$$



But $ cos(infty) $ does not exist right? Does it mean the integral actually diverse?



Edit: The question comes from https://math.uchicago.edu/~min/GRE/files/week1.pdf



Who can point out my mistake in the above deduction?










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  • 2




    Use Laplace transform or add an artificial term $e^{-tx}$ in your integral.
    – Nosrati
    Oct 22 '18 at 13:40








  • 2




    Hi begin{align} int_{0}^{infty}e^{-tx}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}e^{-tx}sin (yx)dy\ &=int_{alpha}^{beta}dyint_{0}^{infty}e^{-tx}sin(yx)dx\ end{align}
    – Nosrati
    Oct 22 '18 at 13:46
















8















Suppose $ alpha, beta>0 $. Compute:
$$ int_{0}^{infty}frac{cos (alpha x)-cos (beta x)}{x}dx $$




Here is what I do:
$$begin{align}
int_{0}^{infty}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}sin (yx)dy\
&=int_{alpha}^{beta}dyint_{0}^{infty}sin(yx)dx\
& \
& qquadtext{let $ yx=u $}\
& \
&=int_{alpha}^{beta}frac{1}{y}dyint_0^{infty}sin u du\
&=int_{alpha}^{beta}frac{1}{y}dyleft( -cos u|_{infty}+cos u|_0 right)\
&=logfrac{beta}{alpha}(-cos(infty)+1)\
&=logfrac{beta}{alpha}-cos(infty)logfrac{beta}{alpha}
end{align}$$



But $ cos(infty) $ does not exist right? Does it mean the integral actually diverse?



Edit: The question comes from https://math.uchicago.edu/~min/GRE/files/week1.pdf



Who can point out my mistake in the above deduction?










share|cite|improve this question




















  • 2




    Use Laplace transform or add an artificial term $e^{-tx}$ in your integral.
    – Nosrati
    Oct 22 '18 at 13:40








  • 2




    Hi begin{align} int_{0}^{infty}e^{-tx}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}e^{-tx}sin (yx)dy\ &=int_{alpha}^{beta}dyint_{0}^{infty}e^{-tx}sin(yx)dx\ end{align}
    – Nosrati
    Oct 22 '18 at 13:46














8












8








8


2






Suppose $ alpha, beta>0 $. Compute:
$$ int_{0}^{infty}frac{cos (alpha x)-cos (beta x)}{x}dx $$




Here is what I do:
$$begin{align}
int_{0}^{infty}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}sin (yx)dy\
&=int_{alpha}^{beta}dyint_{0}^{infty}sin(yx)dx\
& \
& qquadtext{let $ yx=u $}\
& \
&=int_{alpha}^{beta}frac{1}{y}dyint_0^{infty}sin u du\
&=int_{alpha}^{beta}frac{1}{y}dyleft( -cos u|_{infty}+cos u|_0 right)\
&=logfrac{beta}{alpha}(-cos(infty)+1)\
&=logfrac{beta}{alpha}-cos(infty)logfrac{beta}{alpha}
end{align}$$



But $ cos(infty) $ does not exist right? Does it mean the integral actually diverse?



Edit: The question comes from https://math.uchicago.edu/~min/GRE/files/week1.pdf



Who can point out my mistake in the above deduction?










share|cite|improve this question
















Suppose $ alpha, beta>0 $. Compute:
$$ int_{0}^{infty}frac{cos (alpha x)-cos (beta x)}{x}dx $$




Here is what I do:
$$begin{align}
int_{0}^{infty}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}sin (yx)dy\
&=int_{alpha}^{beta}dyint_{0}^{infty}sin(yx)dx\
& \
& qquadtext{let $ yx=u $}\
& \
&=int_{alpha}^{beta}frac{1}{y}dyint_0^{infty}sin u du\
&=int_{alpha}^{beta}frac{1}{y}dyleft( -cos u|_{infty}+cos u|_0 right)\
&=logfrac{beta}{alpha}(-cos(infty)+1)\
&=logfrac{beta}{alpha}-cos(infty)logfrac{beta}{alpha}
end{align}$$



But $ cos(infty) $ does not exist right? Does it mean the integral actually diverse?



Edit: The question comes from https://math.uchicago.edu/~min/GRE/files/week1.pdf



Who can point out my mistake in the above deduction?







calculus integration






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edited Oct 22 '18 at 14:26







Philip

















asked Oct 22 '18 at 13:30









PhilipPhilip

1,0721315




1,0721315








  • 2




    Use Laplace transform or add an artificial term $e^{-tx}$ in your integral.
    – Nosrati
    Oct 22 '18 at 13:40








  • 2




    Hi begin{align} int_{0}^{infty}e^{-tx}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}e^{-tx}sin (yx)dy\ &=int_{alpha}^{beta}dyint_{0}^{infty}e^{-tx}sin(yx)dx\ end{align}
    – Nosrati
    Oct 22 '18 at 13:46














  • 2




    Use Laplace transform or add an artificial term $e^{-tx}$ in your integral.
    – Nosrati
    Oct 22 '18 at 13:40








  • 2




    Hi begin{align} int_{0}^{infty}e^{-tx}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}e^{-tx}sin (yx)dy\ &=int_{alpha}^{beta}dyint_{0}^{infty}e^{-tx}sin(yx)dx\ end{align}
    – Nosrati
    Oct 22 '18 at 13:46








2




2




Use Laplace transform or add an artificial term $e^{-tx}$ in your integral.
– Nosrati
Oct 22 '18 at 13:40






Use Laplace transform or add an artificial term $e^{-tx}$ in your integral.
– Nosrati
Oct 22 '18 at 13:40






2




2




Hi begin{align} int_{0}^{infty}e^{-tx}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}e^{-tx}sin (yx)dy\ &=int_{alpha}^{beta}dyint_{0}^{infty}e^{-tx}sin(yx)dx\ end{align}
– Nosrati
Oct 22 '18 at 13:46




Hi begin{align} int_{0}^{infty}e^{-tx}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}e^{-tx}sin (yx)dy\ &=int_{alpha}^{beta}dyint_{0}^{infty}e^{-tx}sin(yx)dx\ end{align}
– Nosrati
Oct 22 '18 at 13:46










3 Answers
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8














begin{align}
int_{0}^{infty}e^{-tx}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}e^{-tx}sin (yx)dy\
&=int_{alpha}^{beta}dyint_{0}^{infty}e^{-tx}sin(yx)dx\
&=int_{alpha}^{beta}dydfrac{y}{t^2+y^2}\
&=dfrac12lndfrac{t^2+beta^2}{t^2+alpha^2}
end{align}

now let $t=0$.






share|cite|improve this answer

















  • 1




    You need to justify your interchanging the order of integration.
    – Mark Viola
    Oct 22 '18 at 14:50



















6














This is a Frullani integral. Compute as follows.



Let $0 < r < R < +infty$. Then
$$newcommand diff {,mathrm d}
int_r^R frac {cos(alpha x) - cos(beta x)}x diff x = int_r^R frac {cos(alpha x)}x diff x - int_r^R frac {cos(beta x)}x diff x =left( int_{alpha r}^{alpha R} - int_{beta r}^{beta R}right)frac {cos t}t diff t = int_{alpha r}^{beta r} frac {cos t}tdiff t - int_{alpha R}^{beta R} frac {cos t} t diff t = I(r) - J(R).
$$

Now for $I(r)$, use the 1st MVT for integrals, we have
$$
I(r) = cos(A) int_{alpha r}^{beta r} frac {diff t} t = cos(A) log(beta/alpha) [A = alpha r + (1-s)(beta - alpha)r, s in (0,1)] xrightarrow{r to 0^+} cos 0 log(beta /alpha) = log(beta/alpha).
$$

For $J(R)$, note that the integral
$$
int_1^{+infty}frac {cos t}t diff t
$$

converges by Dirichlet test, hence
$$
J(R) xrightarrow{R to +infty} 0
$$

by Cauchy principle. Altogether the original integral is
$$
lim_{substack {r to 0^+\ Rto +infty }} I(r) - J(R) = logleft( frac beta alpharight).
$$






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  • To whoever edited my post, thanks for altering the name of the person! I am not a native, so the textbook I used might give the wrong spelling. Anyway, thanks!
    – xbh
    Oct 23 '18 at 2:55



















4














I would actually use Laplace transforms to compute this sort of an integral.
You must have used it in solving linear differential equations in the past.
It can be defined as follows:-
$$ mathcal{L}{f(x)}=int_{0}^infty e^{-px}f(x)dx = F(p)$$
Here we are transforming the function $f$ with domain $x$ to a function $F$ with domain $p$ with a unilateral integral transform of kernel $e^{-px}$. Now consider the general Laplace transform formula given above. Differentiating both sides with respect to $p$ we get:
$$F'(p)=int_{0}^infty e^{-px}(-x)f(x)dx=-mathcal{L}{xf(x)} rightarrow (1)$$
Now put $G(p)$ as Laplace transform of $f(x)/x$ and obtain its differentiation using equation $(1)$:-
$$G(p)=mathcal{L}left{frac{f(x)}{x}right}Rightarrow G'(p)=-mathcal{L}{f(x)}=-F(p)rightarrow (2)$$
Using the Fundamental Theorem of Calculus (relationship between derivative and integral) for $(2)$:
$$G(p)=-int_{a}^p F(p)dp Rightarrow int_{0}^infty e^{-px}frac{f(x)}{x}dx=-int_{a}^p F(p)dp rightarrow (3)$$
Note that $a$ here is some constant. If $G(p) rightarrow 0$ as $p rightarrow infty$ then we put $a = infty$ and obtain the following:-
$$int_{0}^infty e^{-px}frac{f(x)}{x}dx=int_{p}^infty F(p)dp rightarrow (4)$$
If we let $p rightarrow 0$ on both sides of equation $(4)$ we get the following:
$$int_{0}^infty frac{f(x)}{x}dx=int_{0}^infty F(p)dp rightarrow (5)$$
This is useful for us for finding the improper integral of various functions of the form $f(x)/x$ where the transform $F(p)$ is known. Now I leave the proof for the following upto you (which is elementary considering we use integration by parts):
$$mathcal{L}{ cos bx } = int_{0}^infty e^{-px}(cos bx) dx = frac{p}{p^2 + b^2} (p>0) rightarrow (6)$$
For some constant $b$. Now using equation (5) and (6) we get:
$$int_{0}^infty frac{cos bx}{x}dx=int_{0}^infty frac{p}{p^2 + b^2}dp rightarrow (7)$$
Now plugging in equation $(7)$ into the integral we are required to compute:-
$$I=int_{0}^infty frac{cos alpha x - cos beta x}{x}dx = int_{0}^infty p left(
frac{1}{p^2 + alpha^2} - frac{1}{p^2 + beta^2} right)dp$$

$$Rightarrow I=frac{beta^2-alpha^2}{2} int_{0}^infty frac{2p}{(p^2+alpha^2)(p^2 +beta^2)}dp rightarrow (8)$$
Set $v=frac{beta^2+alpha^2}{2}; u=frac{beta^2-alpha^2}{2}; t=p^2+u$ and using the substitutions and some further simplification:
$$I=int_{v}^infty frac{u}{t^2-u^2}dt = left[frac{1}{2}ln left|frac{t-u}{t+u}right|right]_{t=v}^{t=infty}=frac{1}{2}ln left|frac{u+v}{u-v}right|$$
Substituting the variables back and rewriting the main equation for $I$ we get:
$$int_{0}^infty frac{cos alpha x - cos beta x}{x}dx = ln frac{beta}{alpha}$$






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    3 Answers
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    8














    begin{align}
    int_{0}^{infty}e^{-tx}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}e^{-tx}sin (yx)dy\
    &=int_{alpha}^{beta}dyint_{0}^{infty}e^{-tx}sin(yx)dx\
    &=int_{alpha}^{beta}dydfrac{y}{t^2+y^2}\
    &=dfrac12lndfrac{t^2+beta^2}{t^2+alpha^2}
    end{align}

    now let $t=0$.






    share|cite|improve this answer

















    • 1




      You need to justify your interchanging the order of integration.
      – Mark Viola
      Oct 22 '18 at 14:50
















    8














    begin{align}
    int_{0}^{infty}e^{-tx}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}e^{-tx}sin (yx)dy\
    &=int_{alpha}^{beta}dyint_{0}^{infty}e^{-tx}sin(yx)dx\
    &=int_{alpha}^{beta}dydfrac{y}{t^2+y^2}\
    &=dfrac12lndfrac{t^2+beta^2}{t^2+alpha^2}
    end{align}

    now let $t=0$.






    share|cite|improve this answer

















    • 1




      You need to justify your interchanging the order of integration.
      – Mark Viola
      Oct 22 '18 at 14:50














    8












    8








    8






    begin{align}
    int_{0}^{infty}e^{-tx}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}e^{-tx}sin (yx)dy\
    &=int_{alpha}^{beta}dyint_{0}^{infty}e^{-tx}sin(yx)dx\
    &=int_{alpha}^{beta}dydfrac{y}{t^2+y^2}\
    &=dfrac12lndfrac{t^2+beta^2}{t^2+alpha^2}
    end{align}

    now let $t=0$.






    share|cite|improve this answer












    begin{align}
    int_{0}^{infty}e^{-tx}frac{cos (alpha x)-cos (beta x)}{x}dx &= int_{0}^{infty}dxint_{alpha}^{beta}e^{-tx}sin (yx)dy\
    &=int_{alpha}^{beta}dyint_{0}^{infty}e^{-tx}sin(yx)dx\
    &=int_{alpha}^{beta}dydfrac{y}{t^2+y^2}\
    &=dfrac12lndfrac{t^2+beta^2}{t^2+alpha^2}
    end{align}

    now let $t=0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 22 '18 at 14:11









    NosratiNosrati

    26.5k62354




    26.5k62354








    • 1




      You need to justify your interchanging the order of integration.
      – Mark Viola
      Oct 22 '18 at 14:50














    • 1




      You need to justify your interchanging the order of integration.
      – Mark Viola
      Oct 22 '18 at 14:50








    1




    1




    You need to justify your interchanging the order of integration.
    – Mark Viola
    Oct 22 '18 at 14:50




    You need to justify your interchanging the order of integration.
    – Mark Viola
    Oct 22 '18 at 14:50











    6














    This is a Frullani integral. Compute as follows.



    Let $0 < r < R < +infty$. Then
    $$newcommand diff {,mathrm d}
    int_r^R frac {cos(alpha x) - cos(beta x)}x diff x = int_r^R frac {cos(alpha x)}x diff x - int_r^R frac {cos(beta x)}x diff x =left( int_{alpha r}^{alpha R} - int_{beta r}^{beta R}right)frac {cos t}t diff t = int_{alpha r}^{beta r} frac {cos t}tdiff t - int_{alpha R}^{beta R} frac {cos t} t diff t = I(r) - J(R).
    $$

    Now for $I(r)$, use the 1st MVT for integrals, we have
    $$
    I(r) = cos(A) int_{alpha r}^{beta r} frac {diff t} t = cos(A) log(beta/alpha) [A = alpha r + (1-s)(beta - alpha)r, s in (0,1)] xrightarrow{r to 0^+} cos 0 log(beta /alpha) = log(beta/alpha).
    $$

    For $J(R)$, note that the integral
    $$
    int_1^{+infty}frac {cos t}t diff t
    $$

    converges by Dirichlet test, hence
    $$
    J(R) xrightarrow{R to +infty} 0
    $$

    by Cauchy principle. Altogether the original integral is
    $$
    lim_{substack {r to 0^+\ Rto +infty }} I(r) - J(R) = logleft( frac beta alpharight).
    $$






    share|cite|improve this answer























    • To whoever edited my post, thanks for altering the name of the person! I am not a native, so the textbook I used might give the wrong spelling. Anyway, thanks!
      – xbh
      Oct 23 '18 at 2:55
















    6














    This is a Frullani integral. Compute as follows.



    Let $0 < r < R < +infty$. Then
    $$newcommand diff {,mathrm d}
    int_r^R frac {cos(alpha x) - cos(beta x)}x diff x = int_r^R frac {cos(alpha x)}x diff x - int_r^R frac {cos(beta x)}x diff x =left( int_{alpha r}^{alpha R} - int_{beta r}^{beta R}right)frac {cos t}t diff t = int_{alpha r}^{beta r} frac {cos t}tdiff t - int_{alpha R}^{beta R} frac {cos t} t diff t = I(r) - J(R).
    $$

    Now for $I(r)$, use the 1st MVT for integrals, we have
    $$
    I(r) = cos(A) int_{alpha r}^{beta r} frac {diff t} t = cos(A) log(beta/alpha) [A = alpha r + (1-s)(beta - alpha)r, s in (0,1)] xrightarrow{r to 0^+} cos 0 log(beta /alpha) = log(beta/alpha).
    $$

    For $J(R)$, note that the integral
    $$
    int_1^{+infty}frac {cos t}t diff t
    $$

    converges by Dirichlet test, hence
    $$
    J(R) xrightarrow{R to +infty} 0
    $$

    by Cauchy principle. Altogether the original integral is
    $$
    lim_{substack {r to 0^+\ Rto +infty }} I(r) - J(R) = logleft( frac beta alpharight).
    $$






    share|cite|improve this answer























    • To whoever edited my post, thanks for altering the name of the person! I am not a native, so the textbook I used might give the wrong spelling. Anyway, thanks!
      – xbh
      Oct 23 '18 at 2:55














    6












    6








    6






    This is a Frullani integral. Compute as follows.



    Let $0 < r < R < +infty$. Then
    $$newcommand diff {,mathrm d}
    int_r^R frac {cos(alpha x) - cos(beta x)}x diff x = int_r^R frac {cos(alpha x)}x diff x - int_r^R frac {cos(beta x)}x diff x =left( int_{alpha r}^{alpha R} - int_{beta r}^{beta R}right)frac {cos t}t diff t = int_{alpha r}^{beta r} frac {cos t}tdiff t - int_{alpha R}^{beta R} frac {cos t} t diff t = I(r) - J(R).
    $$

    Now for $I(r)$, use the 1st MVT for integrals, we have
    $$
    I(r) = cos(A) int_{alpha r}^{beta r} frac {diff t} t = cos(A) log(beta/alpha) [A = alpha r + (1-s)(beta - alpha)r, s in (0,1)] xrightarrow{r to 0^+} cos 0 log(beta /alpha) = log(beta/alpha).
    $$

    For $J(R)$, note that the integral
    $$
    int_1^{+infty}frac {cos t}t diff t
    $$

    converges by Dirichlet test, hence
    $$
    J(R) xrightarrow{R to +infty} 0
    $$

    by Cauchy principle. Altogether the original integral is
    $$
    lim_{substack {r to 0^+\ Rto +infty }} I(r) - J(R) = logleft( frac beta alpharight).
    $$






    share|cite|improve this answer














    This is a Frullani integral. Compute as follows.



    Let $0 < r < R < +infty$. Then
    $$newcommand diff {,mathrm d}
    int_r^R frac {cos(alpha x) - cos(beta x)}x diff x = int_r^R frac {cos(alpha x)}x diff x - int_r^R frac {cos(beta x)}x diff x =left( int_{alpha r}^{alpha R} - int_{beta r}^{beta R}right)frac {cos t}t diff t = int_{alpha r}^{beta r} frac {cos t}tdiff t - int_{alpha R}^{beta R} frac {cos t} t diff t = I(r) - J(R).
    $$

    Now for $I(r)$, use the 1st MVT for integrals, we have
    $$
    I(r) = cos(A) int_{alpha r}^{beta r} frac {diff t} t = cos(A) log(beta/alpha) [A = alpha r + (1-s)(beta - alpha)r, s in (0,1)] xrightarrow{r to 0^+} cos 0 log(beta /alpha) = log(beta/alpha).
    $$

    For $J(R)$, note that the integral
    $$
    int_1^{+infty}frac {cos t}t diff t
    $$

    converges by Dirichlet test, hence
    $$
    J(R) xrightarrow{R to +infty} 0
    $$

    by Cauchy principle. Altogether the original integral is
    $$
    lim_{substack {r to 0^+\ Rto +infty }} I(r) - J(R) = logleft( frac beta alpharight).
    $$







    share|cite|improve this answer














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    edited Oct 22 '18 at 14:48









    Mark Viola

    130k1274170




    130k1274170










    answered Oct 22 '18 at 14:15









    xbhxbh

    5,8281522




    5,8281522












    • To whoever edited my post, thanks for altering the name of the person! I am not a native, so the textbook I used might give the wrong spelling. Anyway, thanks!
      – xbh
      Oct 23 '18 at 2:55


















    • To whoever edited my post, thanks for altering the name of the person! I am not a native, so the textbook I used might give the wrong spelling. Anyway, thanks!
      – xbh
      Oct 23 '18 at 2:55
















    To whoever edited my post, thanks for altering the name of the person! I am not a native, so the textbook I used might give the wrong spelling. Anyway, thanks!
    – xbh
    Oct 23 '18 at 2:55




    To whoever edited my post, thanks for altering the name of the person! I am not a native, so the textbook I used might give the wrong spelling. Anyway, thanks!
    – xbh
    Oct 23 '18 at 2:55











    4














    I would actually use Laplace transforms to compute this sort of an integral.
    You must have used it in solving linear differential equations in the past.
    It can be defined as follows:-
    $$ mathcal{L}{f(x)}=int_{0}^infty e^{-px}f(x)dx = F(p)$$
    Here we are transforming the function $f$ with domain $x$ to a function $F$ with domain $p$ with a unilateral integral transform of kernel $e^{-px}$. Now consider the general Laplace transform formula given above. Differentiating both sides with respect to $p$ we get:
    $$F'(p)=int_{0}^infty e^{-px}(-x)f(x)dx=-mathcal{L}{xf(x)} rightarrow (1)$$
    Now put $G(p)$ as Laplace transform of $f(x)/x$ and obtain its differentiation using equation $(1)$:-
    $$G(p)=mathcal{L}left{frac{f(x)}{x}right}Rightarrow G'(p)=-mathcal{L}{f(x)}=-F(p)rightarrow (2)$$
    Using the Fundamental Theorem of Calculus (relationship between derivative and integral) for $(2)$:
    $$G(p)=-int_{a}^p F(p)dp Rightarrow int_{0}^infty e^{-px}frac{f(x)}{x}dx=-int_{a}^p F(p)dp rightarrow (3)$$
    Note that $a$ here is some constant. If $G(p) rightarrow 0$ as $p rightarrow infty$ then we put $a = infty$ and obtain the following:-
    $$int_{0}^infty e^{-px}frac{f(x)}{x}dx=int_{p}^infty F(p)dp rightarrow (4)$$
    If we let $p rightarrow 0$ on both sides of equation $(4)$ we get the following:
    $$int_{0}^infty frac{f(x)}{x}dx=int_{0}^infty F(p)dp rightarrow (5)$$
    This is useful for us for finding the improper integral of various functions of the form $f(x)/x$ where the transform $F(p)$ is known. Now I leave the proof for the following upto you (which is elementary considering we use integration by parts):
    $$mathcal{L}{ cos bx } = int_{0}^infty e^{-px}(cos bx) dx = frac{p}{p^2 + b^2} (p>0) rightarrow (6)$$
    For some constant $b$. Now using equation (5) and (6) we get:
    $$int_{0}^infty frac{cos bx}{x}dx=int_{0}^infty frac{p}{p^2 + b^2}dp rightarrow (7)$$
    Now plugging in equation $(7)$ into the integral we are required to compute:-
    $$I=int_{0}^infty frac{cos alpha x - cos beta x}{x}dx = int_{0}^infty p left(
    frac{1}{p^2 + alpha^2} - frac{1}{p^2 + beta^2} right)dp$$

    $$Rightarrow I=frac{beta^2-alpha^2}{2} int_{0}^infty frac{2p}{(p^2+alpha^2)(p^2 +beta^2)}dp rightarrow (8)$$
    Set $v=frac{beta^2+alpha^2}{2}; u=frac{beta^2-alpha^2}{2}; t=p^2+u$ and using the substitutions and some further simplification:
    $$I=int_{v}^infty frac{u}{t^2-u^2}dt = left[frac{1}{2}ln left|frac{t-u}{t+u}right|right]_{t=v}^{t=infty}=frac{1}{2}ln left|frac{u+v}{u-v}right|$$
    Substituting the variables back and rewriting the main equation for $I$ we get:
    $$int_{0}^infty frac{cos alpha x - cos beta x}{x}dx = ln frac{beta}{alpha}$$






    share|cite|improve this answer




























      4














      I would actually use Laplace transforms to compute this sort of an integral.
      You must have used it in solving linear differential equations in the past.
      It can be defined as follows:-
      $$ mathcal{L}{f(x)}=int_{0}^infty e^{-px}f(x)dx = F(p)$$
      Here we are transforming the function $f$ with domain $x$ to a function $F$ with domain $p$ with a unilateral integral transform of kernel $e^{-px}$. Now consider the general Laplace transform formula given above. Differentiating both sides with respect to $p$ we get:
      $$F'(p)=int_{0}^infty e^{-px}(-x)f(x)dx=-mathcal{L}{xf(x)} rightarrow (1)$$
      Now put $G(p)$ as Laplace transform of $f(x)/x$ and obtain its differentiation using equation $(1)$:-
      $$G(p)=mathcal{L}left{frac{f(x)}{x}right}Rightarrow G'(p)=-mathcal{L}{f(x)}=-F(p)rightarrow (2)$$
      Using the Fundamental Theorem of Calculus (relationship between derivative and integral) for $(2)$:
      $$G(p)=-int_{a}^p F(p)dp Rightarrow int_{0}^infty e^{-px}frac{f(x)}{x}dx=-int_{a}^p F(p)dp rightarrow (3)$$
      Note that $a$ here is some constant. If $G(p) rightarrow 0$ as $p rightarrow infty$ then we put $a = infty$ and obtain the following:-
      $$int_{0}^infty e^{-px}frac{f(x)}{x}dx=int_{p}^infty F(p)dp rightarrow (4)$$
      If we let $p rightarrow 0$ on both sides of equation $(4)$ we get the following:
      $$int_{0}^infty frac{f(x)}{x}dx=int_{0}^infty F(p)dp rightarrow (5)$$
      This is useful for us for finding the improper integral of various functions of the form $f(x)/x$ where the transform $F(p)$ is known. Now I leave the proof for the following upto you (which is elementary considering we use integration by parts):
      $$mathcal{L}{ cos bx } = int_{0}^infty e^{-px}(cos bx) dx = frac{p}{p^2 + b^2} (p>0) rightarrow (6)$$
      For some constant $b$. Now using equation (5) and (6) we get:
      $$int_{0}^infty frac{cos bx}{x}dx=int_{0}^infty frac{p}{p^2 + b^2}dp rightarrow (7)$$
      Now plugging in equation $(7)$ into the integral we are required to compute:-
      $$I=int_{0}^infty frac{cos alpha x - cos beta x}{x}dx = int_{0}^infty p left(
      frac{1}{p^2 + alpha^2} - frac{1}{p^2 + beta^2} right)dp$$

      $$Rightarrow I=frac{beta^2-alpha^2}{2} int_{0}^infty frac{2p}{(p^2+alpha^2)(p^2 +beta^2)}dp rightarrow (8)$$
      Set $v=frac{beta^2+alpha^2}{2}; u=frac{beta^2-alpha^2}{2}; t=p^2+u$ and using the substitutions and some further simplification:
      $$I=int_{v}^infty frac{u}{t^2-u^2}dt = left[frac{1}{2}ln left|frac{t-u}{t+u}right|right]_{t=v}^{t=infty}=frac{1}{2}ln left|frac{u+v}{u-v}right|$$
      Substituting the variables back and rewriting the main equation for $I$ we get:
      $$int_{0}^infty frac{cos alpha x - cos beta x}{x}dx = ln frac{beta}{alpha}$$






      share|cite|improve this answer


























        4












        4








        4






        I would actually use Laplace transforms to compute this sort of an integral.
        You must have used it in solving linear differential equations in the past.
        It can be defined as follows:-
        $$ mathcal{L}{f(x)}=int_{0}^infty e^{-px}f(x)dx = F(p)$$
        Here we are transforming the function $f$ with domain $x$ to a function $F$ with domain $p$ with a unilateral integral transform of kernel $e^{-px}$. Now consider the general Laplace transform formula given above. Differentiating both sides with respect to $p$ we get:
        $$F'(p)=int_{0}^infty e^{-px}(-x)f(x)dx=-mathcal{L}{xf(x)} rightarrow (1)$$
        Now put $G(p)$ as Laplace transform of $f(x)/x$ and obtain its differentiation using equation $(1)$:-
        $$G(p)=mathcal{L}left{frac{f(x)}{x}right}Rightarrow G'(p)=-mathcal{L}{f(x)}=-F(p)rightarrow (2)$$
        Using the Fundamental Theorem of Calculus (relationship between derivative and integral) for $(2)$:
        $$G(p)=-int_{a}^p F(p)dp Rightarrow int_{0}^infty e^{-px}frac{f(x)}{x}dx=-int_{a}^p F(p)dp rightarrow (3)$$
        Note that $a$ here is some constant. If $G(p) rightarrow 0$ as $p rightarrow infty$ then we put $a = infty$ and obtain the following:-
        $$int_{0}^infty e^{-px}frac{f(x)}{x}dx=int_{p}^infty F(p)dp rightarrow (4)$$
        If we let $p rightarrow 0$ on both sides of equation $(4)$ we get the following:
        $$int_{0}^infty frac{f(x)}{x}dx=int_{0}^infty F(p)dp rightarrow (5)$$
        This is useful for us for finding the improper integral of various functions of the form $f(x)/x$ where the transform $F(p)$ is known. Now I leave the proof for the following upto you (which is elementary considering we use integration by parts):
        $$mathcal{L}{ cos bx } = int_{0}^infty e^{-px}(cos bx) dx = frac{p}{p^2 + b^2} (p>0) rightarrow (6)$$
        For some constant $b$. Now using equation (5) and (6) we get:
        $$int_{0}^infty frac{cos bx}{x}dx=int_{0}^infty frac{p}{p^2 + b^2}dp rightarrow (7)$$
        Now plugging in equation $(7)$ into the integral we are required to compute:-
        $$I=int_{0}^infty frac{cos alpha x - cos beta x}{x}dx = int_{0}^infty p left(
        frac{1}{p^2 + alpha^2} - frac{1}{p^2 + beta^2} right)dp$$

        $$Rightarrow I=frac{beta^2-alpha^2}{2} int_{0}^infty frac{2p}{(p^2+alpha^2)(p^2 +beta^2)}dp rightarrow (8)$$
        Set $v=frac{beta^2+alpha^2}{2}; u=frac{beta^2-alpha^2}{2}; t=p^2+u$ and using the substitutions and some further simplification:
        $$I=int_{v}^infty frac{u}{t^2-u^2}dt = left[frac{1}{2}ln left|frac{t-u}{t+u}right|right]_{t=v}^{t=infty}=frac{1}{2}ln left|frac{u+v}{u-v}right|$$
        Substituting the variables back and rewriting the main equation for $I$ we get:
        $$int_{0}^infty frac{cos alpha x - cos beta x}{x}dx = ln frac{beta}{alpha}$$






        share|cite|improve this answer














        I would actually use Laplace transforms to compute this sort of an integral.
        You must have used it in solving linear differential equations in the past.
        It can be defined as follows:-
        $$ mathcal{L}{f(x)}=int_{0}^infty e^{-px}f(x)dx = F(p)$$
        Here we are transforming the function $f$ with domain $x$ to a function $F$ with domain $p$ with a unilateral integral transform of kernel $e^{-px}$. Now consider the general Laplace transform formula given above. Differentiating both sides with respect to $p$ we get:
        $$F'(p)=int_{0}^infty e^{-px}(-x)f(x)dx=-mathcal{L}{xf(x)} rightarrow (1)$$
        Now put $G(p)$ as Laplace transform of $f(x)/x$ and obtain its differentiation using equation $(1)$:-
        $$G(p)=mathcal{L}left{frac{f(x)}{x}right}Rightarrow G'(p)=-mathcal{L}{f(x)}=-F(p)rightarrow (2)$$
        Using the Fundamental Theorem of Calculus (relationship between derivative and integral) for $(2)$:
        $$G(p)=-int_{a}^p F(p)dp Rightarrow int_{0}^infty e^{-px}frac{f(x)}{x}dx=-int_{a}^p F(p)dp rightarrow (3)$$
        Note that $a$ here is some constant. If $G(p) rightarrow 0$ as $p rightarrow infty$ then we put $a = infty$ and obtain the following:-
        $$int_{0}^infty e^{-px}frac{f(x)}{x}dx=int_{p}^infty F(p)dp rightarrow (4)$$
        If we let $p rightarrow 0$ on both sides of equation $(4)$ we get the following:
        $$int_{0}^infty frac{f(x)}{x}dx=int_{0}^infty F(p)dp rightarrow (5)$$
        This is useful for us for finding the improper integral of various functions of the form $f(x)/x$ where the transform $F(p)$ is known. Now I leave the proof for the following upto you (which is elementary considering we use integration by parts):
        $$mathcal{L}{ cos bx } = int_{0}^infty e^{-px}(cos bx) dx = frac{p}{p^2 + b^2} (p>0) rightarrow (6)$$
        For some constant $b$. Now using equation (5) and (6) we get:
        $$int_{0}^infty frac{cos bx}{x}dx=int_{0}^infty frac{p}{p^2 + b^2}dp rightarrow (7)$$
        Now plugging in equation $(7)$ into the integral we are required to compute:-
        $$I=int_{0}^infty frac{cos alpha x - cos beta x}{x}dx = int_{0}^infty p left(
        frac{1}{p^2 + alpha^2} - frac{1}{p^2 + beta^2} right)dp$$

        $$Rightarrow I=frac{beta^2-alpha^2}{2} int_{0}^infty frac{2p}{(p^2+alpha^2)(p^2 +beta^2)}dp rightarrow (8)$$
        Set $v=frac{beta^2+alpha^2}{2}; u=frac{beta^2-alpha^2}{2}; t=p^2+u$ and using the substitutions and some further simplification:
        $$I=int_{v}^infty frac{u}{t^2-u^2}dt = left[frac{1}{2}ln left|frac{t-u}{t+u}right|right]_{t=v}^{t=infty}=frac{1}{2}ln left|frac{u+v}{u-v}right|$$
        Substituting the variables back and rewriting the main equation for $I$ we get:
        $$int_{0}^infty frac{cos alpha x - cos beta x}{x}dx = ln frac{beta}{alpha}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 14:10

























        answered Oct 22 '18 at 16:21









        Suhrid SahaSuhrid Saha

        1307




        1307






























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