$I_{m,n}=intfrac{x^m}{(ax^2+bx+c)^n}dx$ Reduction Formula
$begingroup$
I'm having trouble proving the following reduction formula:
If
$$I_{m,n}=intfrac{x^m}{(ax^2+bx+c)^n}dx$$
then
$$intfrac{x^m}{(ax^2+bx+c)^n}dx=-frac{x^{m-1}}{a(2m-n-1)(ax^2+bx+c)^{n-1}}-frac{b(n-m)}{a(2m-n-1)}I_{m-1,n}+frac{c(m-1)}{a(2m-n-1)}I_{m-2,n}$$
My attempt went as follows:
$$intfrac{x^m}{(ax^2+bx+c)^n}dxspacebegin{vmatrix}u=frac{1}{(ax^2+bx+c)^n}\du=frac{-n(2ax+b)}{(ax^2+bx+c)^{n+1}}dxend{vmatrix}space dv=x^mdxquad v=frac{1}{m+1}x^{m+1}\intfrac{x^m}{(ax^2+bx+c)^n}dx=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)intfrac{(2ax+b)x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)bigg(2aintfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bintfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dxbigg)\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{2an}{m+1}bigg)intfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bigg(frac{bn}{m+1}bigg)intfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{1}{m+1}I_{m+1,n}+bigg(frac{2an}{m+1}bigg)I_{m+2,n+1}+bigg(frac{bn}{m+1}bigg)I_{m+1,n+1}$$
This is where I'm stuck, any help?
integration reduction-formula
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add a comment |
$begingroup$
I'm having trouble proving the following reduction formula:
If
$$I_{m,n}=intfrac{x^m}{(ax^2+bx+c)^n}dx$$
then
$$intfrac{x^m}{(ax^2+bx+c)^n}dx=-frac{x^{m-1}}{a(2m-n-1)(ax^2+bx+c)^{n-1}}-frac{b(n-m)}{a(2m-n-1)}I_{m-1,n}+frac{c(m-1)}{a(2m-n-1)}I_{m-2,n}$$
My attempt went as follows:
$$intfrac{x^m}{(ax^2+bx+c)^n}dxspacebegin{vmatrix}u=frac{1}{(ax^2+bx+c)^n}\du=frac{-n(2ax+b)}{(ax^2+bx+c)^{n+1}}dxend{vmatrix}space dv=x^mdxquad v=frac{1}{m+1}x^{m+1}\intfrac{x^m}{(ax^2+bx+c)^n}dx=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)intfrac{(2ax+b)x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)bigg(2aintfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bintfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dxbigg)\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{2an}{m+1}bigg)intfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bigg(frac{bn}{m+1}bigg)intfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{1}{m+1}I_{m+1,n}+bigg(frac{2an}{m+1}bigg)I_{m+2,n+1}+bigg(frac{bn}{m+1}bigg)I_{m+1,n+1}$$
This is where I'm stuck, any help?
integration reduction-formula
$endgroup$
$begingroup$
What are the restrictions on $m$ and $n$?
$endgroup$
– clathratus
Jan 8 at 2:49
$begingroup$
Also is the $I_{m+1,n}$ in the last line is a typo?
$endgroup$
– clathratus
Jan 8 at 2:50
add a comment |
$begingroup$
I'm having trouble proving the following reduction formula:
If
$$I_{m,n}=intfrac{x^m}{(ax^2+bx+c)^n}dx$$
then
$$intfrac{x^m}{(ax^2+bx+c)^n}dx=-frac{x^{m-1}}{a(2m-n-1)(ax^2+bx+c)^{n-1}}-frac{b(n-m)}{a(2m-n-1)}I_{m-1,n}+frac{c(m-1)}{a(2m-n-1)}I_{m-2,n}$$
My attempt went as follows:
$$intfrac{x^m}{(ax^2+bx+c)^n}dxspacebegin{vmatrix}u=frac{1}{(ax^2+bx+c)^n}\du=frac{-n(2ax+b)}{(ax^2+bx+c)^{n+1}}dxend{vmatrix}space dv=x^mdxquad v=frac{1}{m+1}x^{m+1}\intfrac{x^m}{(ax^2+bx+c)^n}dx=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)intfrac{(2ax+b)x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)bigg(2aintfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bintfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dxbigg)\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{2an}{m+1}bigg)intfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bigg(frac{bn}{m+1}bigg)intfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{1}{m+1}I_{m+1,n}+bigg(frac{2an}{m+1}bigg)I_{m+2,n+1}+bigg(frac{bn}{m+1}bigg)I_{m+1,n+1}$$
This is where I'm stuck, any help?
integration reduction-formula
$endgroup$
I'm having trouble proving the following reduction formula:
If
$$I_{m,n}=intfrac{x^m}{(ax^2+bx+c)^n}dx$$
then
$$intfrac{x^m}{(ax^2+bx+c)^n}dx=-frac{x^{m-1}}{a(2m-n-1)(ax^2+bx+c)^{n-1}}-frac{b(n-m)}{a(2m-n-1)}I_{m-1,n}+frac{c(m-1)}{a(2m-n-1)}I_{m-2,n}$$
My attempt went as follows:
$$intfrac{x^m}{(ax^2+bx+c)^n}dxspacebegin{vmatrix}u=frac{1}{(ax^2+bx+c)^n}\du=frac{-n(2ax+b)}{(ax^2+bx+c)^{n+1}}dxend{vmatrix}space dv=x^mdxquad v=frac{1}{m+1}x^{m+1}\intfrac{x^m}{(ax^2+bx+c)^n}dx=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)intfrac{(2ax+b)x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)bigg(2aintfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bintfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dxbigg)\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{2an}{m+1}bigg)intfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bigg(frac{bn}{m+1}bigg)intfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{1}{m+1}I_{m+1,n}+bigg(frac{2an}{m+1}bigg)I_{m+2,n+1}+bigg(frac{bn}{m+1}bigg)I_{m+1,n+1}$$
This is where I'm stuck, any help?
integration reduction-formula
integration reduction-formula
edited Jan 8 at 2:37
clathratus
3,740333
3,740333
asked Jan 5 at 23:16
Anson PangAnson Pang
6215
6215
$begingroup$
What are the restrictions on $m$ and $n$?
$endgroup$
– clathratus
Jan 8 at 2:49
$begingroup$
Also is the $I_{m+1,n}$ in the last line is a typo?
$endgroup$
– clathratus
Jan 8 at 2:50
add a comment |
$begingroup$
What are the restrictions on $m$ and $n$?
$endgroup$
– clathratus
Jan 8 at 2:49
$begingroup$
Also is the $I_{m+1,n}$ in the last line is a typo?
$endgroup$
– clathratus
Jan 8 at 2:50
$begingroup$
What are the restrictions on $m$ and $n$?
$endgroup$
– clathratus
Jan 8 at 2:49
$begingroup$
What are the restrictions on $m$ and $n$?
$endgroup$
– clathratus
Jan 8 at 2:49
$begingroup$
Also is the $I_{m+1,n}$ in the last line is a typo?
$endgroup$
– clathratus
Jan 8 at 2:50
$begingroup$
Also is the $I_{m+1,n}$ in the last line is a typo?
$endgroup$
– clathratus
Jan 8 at 2:50
add a comment |
1 Answer
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oldest
votes
$begingroup$
So this means you are using the integration by parts in the wrong direction or with a wrong choice. In the reduction formula you are trying to achieve, you can notice that $m$ is decreasing while $n$ stays constant. Thus, you need a choice of $u,v$ such that:
$$udv = dfrac{x^mdx}{f^n(x)},$$
where $f(x) = ax^2+bx+c$ and:
$$vdu = Cdfrac{x^{m-1}dx}{f^n(x)},$$
with $C$ some constant. If you tried to naively take $u = x^m,$ then you will have $v = intdfrac{dx}{f^n(x)},$ which you cannot evaluate. Therefore, one needs to improvise somehow. For instance,
$$dleft(dfrac{1}{g^k(x)}right) = dfrac{kg'(x)g^{k-1}(x)dx}{g^{2k}(x)} = kdfrac{g'(x)dx}{g^{k+1}(x)}$$ for appropriately smooth function $g(x).$ This is where I will leave you with a hint:
$$dfrac{x^mdx}{f^n(x)} = dfrac{x^{m-1}}{2a}cdotdfrac{d(ax^2+bx+c)}{(ax^2+bx+c)^n} -dfrac{b}{2a}cdotdfrac{x^{m-1}}{(ax^2+bx+c)^n}=dots $$
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add a comment |
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$begingroup$
So this means you are using the integration by parts in the wrong direction or with a wrong choice. In the reduction formula you are trying to achieve, you can notice that $m$ is decreasing while $n$ stays constant. Thus, you need a choice of $u,v$ such that:
$$udv = dfrac{x^mdx}{f^n(x)},$$
where $f(x) = ax^2+bx+c$ and:
$$vdu = Cdfrac{x^{m-1}dx}{f^n(x)},$$
with $C$ some constant. If you tried to naively take $u = x^m,$ then you will have $v = intdfrac{dx}{f^n(x)},$ which you cannot evaluate. Therefore, one needs to improvise somehow. For instance,
$$dleft(dfrac{1}{g^k(x)}right) = dfrac{kg'(x)g^{k-1}(x)dx}{g^{2k}(x)} = kdfrac{g'(x)dx}{g^{k+1}(x)}$$ for appropriately smooth function $g(x).$ This is where I will leave you with a hint:
$$dfrac{x^mdx}{f^n(x)} = dfrac{x^{m-1}}{2a}cdotdfrac{d(ax^2+bx+c)}{(ax^2+bx+c)^n} -dfrac{b}{2a}cdotdfrac{x^{m-1}}{(ax^2+bx+c)^n}=dots $$
$endgroup$
add a comment |
$begingroup$
So this means you are using the integration by parts in the wrong direction or with a wrong choice. In the reduction formula you are trying to achieve, you can notice that $m$ is decreasing while $n$ stays constant. Thus, you need a choice of $u,v$ such that:
$$udv = dfrac{x^mdx}{f^n(x)},$$
where $f(x) = ax^2+bx+c$ and:
$$vdu = Cdfrac{x^{m-1}dx}{f^n(x)},$$
with $C$ some constant. If you tried to naively take $u = x^m,$ then you will have $v = intdfrac{dx}{f^n(x)},$ which you cannot evaluate. Therefore, one needs to improvise somehow. For instance,
$$dleft(dfrac{1}{g^k(x)}right) = dfrac{kg'(x)g^{k-1}(x)dx}{g^{2k}(x)} = kdfrac{g'(x)dx}{g^{k+1}(x)}$$ for appropriately smooth function $g(x).$ This is where I will leave you with a hint:
$$dfrac{x^mdx}{f^n(x)} = dfrac{x^{m-1}}{2a}cdotdfrac{d(ax^2+bx+c)}{(ax^2+bx+c)^n} -dfrac{b}{2a}cdotdfrac{x^{m-1}}{(ax^2+bx+c)^n}=dots $$
$endgroup$
add a comment |
$begingroup$
So this means you are using the integration by parts in the wrong direction or with a wrong choice. In the reduction formula you are trying to achieve, you can notice that $m$ is decreasing while $n$ stays constant. Thus, you need a choice of $u,v$ such that:
$$udv = dfrac{x^mdx}{f^n(x)},$$
where $f(x) = ax^2+bx+c$ and:
$$vdu = Cdfrac{x^{m-1}dx}{f^n(x)},$$
with $C$ some constant. If you tried to naively take $u = x^m,$ then you will have $v = intdfrac{dx}{f^n(x)},$ which you cannot evaluate. Therefore, one needs to improvise somehow. For instance,
$$dleft(dfrac{1}{g^k(x)}right) = dfrac{kg'(x)g^{k-1}(x)dx}{g^{2k}(x)} = kdfrac{g'(x)dx}{g^{k+1}(x)}$$ for appropriately smooth function $g(x).$ This is where I will leave you with a hint:
$$dfrac{x^mdx}{f^n(x)} = dfrac{x^{m-1}}{2a}cdotdfrac{d(ax^2+bx+c)}{(ax^2+bx+c)^n} -dfrac{b}{2a}cdotdfrac{x^{m-1}}{(ax^2+bx+c)^n}=dots $$
$endgroup$
So this means you are using the integration by parts in the wrong direction or with a wrong choice. In the reduction formula you are trying to achieve, you can notice that $m$ is decreasing while $n$ stays constant. Thus, you need a choice of $u,v$ such that:
$$udv = dfrac{x^mdx}{f^n(x)},$$
where $f(x) = ax^2+bx+c$ and:
$$vdu = Cdfrac{x^{m-1}dx}{f^n(x)},$$
with $C$ some constant. If you tried to naively take $u = x^m,$ then you will have $v = intdfrac{dx}{f^n(x)},$ which you cannot evaluate. Therefore, one needs to improvise somehow. For instance,
$$dleft(dfrac{1}{g^k(x)}right) = dfrac{kg'(x)g^{k-1}(x)dx}{g^{2k}(x)} = kdfrac{g'(x)dx}{g^{k+1}(x)}$$ for appropriately smooth function $g(x).$ This is where I will leave you with a hint:
$$dfrac{x^mdx}{f^n(x)} = dfrac{x^{m-1}}{2a}cdotdfrac{d(ax^2+bx+c)}{(ax^2+bx+c)^n} -dfrac{b}{2a}cdotdfrac{x^{m-1}}{(ax^2+bx+c)^n}=dots $$
answered Jan 8 at 2:59
dezdichadodezdichado
6,3761929
6,3761929
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$begingroup$
What are the restrictions on $m$ and $n$?
$endgroup$
– clathratus
Jan 8 at 2:49
$begingroup$
Also is the $I_{m+1,n}$ in the last line is a typo?
$endgroup$
– clathratus
Jan 8 at 2:50