I am trying to evaluate the following integral by parts, but no answers so far.$$\int_{0}^{\infty}\left[\frac{x}{(1+ax)(1+x)}\right]^{K}dx,$$ where $K\in\mathbb{N}$ and $a>0$ is a constant. Can anyone help me?
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Mathematica 9.01 produces $$a^{-K} Gamma[-1 + K] Gamma[1 + K] Hypergeometric2F1Regularized[-1 + K, K, 2 K, (-1 + a)/a].$$ – user64494 Jul 23 '13 at 11:07
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2The integrand can be rewritten as $$ \left( {\frac {1}{ \left( a-1 \right) \left( 1+x \right) }}-{\frac { 1}{ \left( a-1 \right) \left( ax+1 \right) }} \right) ^{K}. $$ Then try to apply the binomial theorem. – user64494 Jul 23 '13 at 11:16
2 Answers
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If $a=1$ then the integral becomes $$I=\int_{0}^{\infty}\frac{x^K}{(1+x)^{2K}}dx$$ Letting $\displaystyle \frac{1}{1+x}=z$ we get $$I=\int_{0}^{1}z^{K-2}(1-z)^K dz=\beta\left(K-1,K+1\right)$$
I'll write down the results for $a\ne 1$ shortly.
Samrat Mukhopadhyay
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Here is a formula in terms of the associated Legendre polynomials
$$\frac{ 2\,{4}^{k-1}\Gamma\left( \frac{1}{2}+k \right)\beta \left( k+1,k-1 \right)}{ {a}^{3/4} \left( a-1 \right) ^{k-1/2} } P^{\frac{1}{2}}_{\frac{1}{2}-k}\left( {\frac {a+1}{2\sqrt {a}}}\right)\quad \left\{k\geq 2 \cap k\in \mathbb{N}\right\},$$
where $\beta(u,v)$ is the beta function.
added: For the case $a=1$, we can find the answer $$ {\frac {2\,{4}^{-k}\,\sqrt {\pi }\,\Gamma \left( k+1 \right) }{ \left( k- 1 \right) \Gamma \left( k+1/2 \right) }} \quad \left\{k\geq 2 \cap k\in \mathbb{N}\right\},$$
by plugging $a=1$ in the integral using this technique.
Mhenni Benghorbal
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@RonGordon: It is clear that it is a special case and it needs to be handled separately by plugging in the integral $a=1$. – Mhenni Benghorbal Jul 23 '13 at 13:10
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I don't mean to be giving you a hard time, but should the result not be continuous in $a$? I'm sure that it works out, but I do not see how this is obvious. – Ron Gordon Jul 23 '13 at 13:47
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2(+1) Just a small comment to make the formula less misterious. Using the change of variables $x=\frac{t}{1-t}$ the integral transforms into $$\int_0^1 \frac{ t^{k} (1-t)^{k-2} }{\left[1-(1-a)t\right]^k} dt.$$ This is clearly a hypergeometric function $_2F_1$ (which can reduce to something else because of special parameter values). The result is continuous in $a$ and it is clear that in the case $a=1$ it can be expressed in terms of gamma functions. – Start wearing purple Jul 23 '13 at 13:58
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@O.L.: This was precisely what I was looking for! You should write this up in a separate post. – Ron Gordon Jul 23 '13 at 14:04
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This doesn't seem to agree with Samrat's answer below for the $ a = 1 $ part. His seems correct though. – Jon Claus Jul 24 '13 at 01:22
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(-1): the procedure does not show the distinguishing features for $K\in\mathbb{N}$ . – doraemonpaul Jul 25 '13 at 23:57
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@doraemonpaul: What distinguishing features are you taking about? – Mhenni Benghorbal Jul 26 '13 at 01:36
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@Mhenni Benghorbal: As $K$ is a natural number, $\dfrac{x^K}{(1+ax)^K(1+x)^K}$ can do partial fraction. – doraemonpaul Jul 26 '13 at 02:51
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@doraemonpaul: You can work it out using the technique you suggested. Thanks for the comment. – Mhenni Benghorbal Jul 26 '13 at 10:14