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(I am sorry it's the first time I ask question...)

I don't know how to use mathematical analysis to computer, maybe it need complex analysis to do but I have not learned it.

I just add my solution.

First I think it is a function of "p". So I differentiate it ,than maybe it will be simplified.But it makes no sense...

$\int_{0}^{\infty} \frac{x^{p}}{1+x^{2}} dx$ $p\in(-1,1)$

I don't know the answer .

Thanks a lot.

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    What makes it amazing, out of curiosity? – 2017-01-02
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    add some own thoughts and efforts – 2017-01-02
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    Math.SE click-bait? – 2017-01-02
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    what is your mathematical background? – 2017-01-02
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    undergraduate sophomore@tired – 2017-01-02
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    ok then it will be a lot of effort for you to understand any answer to this question. maybe it would be a good idea to gain some knowledge about Gamma- and Beta functions – 2017-01-02
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    @tired thank you a lot ! I will read your answer earnestly ! – 2017-01-02
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    Such integral already appeared on MSE many times. It is enough to use Euler's Beta function and the reflection formula for the $\Gamma$ function. – 2017-01-02

1 Answers 1

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$$I(p)=\int_{0}^{\infty} \frac{x^{p}}{1+x^{2}} dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\exp\ p t}{\cosh t}\,dt=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cosh\ p t}{\cosh t}\,dt$$

So, set $f(z)=\frac{\cosh pz}{\cosh z}$, take some large $R$ (we will ultimately take $R\to\infty$) and consider the complex rectangular contour $\gamma_R$ with corners at $\pm R,\pm R+i\pi$, oriented counterclockwise.

Consider $\int_{\gamma_R}f(z)\,dz$, and decompose it as $I_1+I_2+I_3+I_4$ as:

  • $I_1$, the integral along the bottom of the rectangle, is $\int_{-R}^Rf(t)\,dt$
  • $I_3$, the integral along the top of the rectangle, from right to left, is $$-\int_{-R}^Rf(t+i\pi)\,dt=\int_{-R}^R\frac{-\cosh(pt+i\pi p)}{\cosh(t+i\pi)}\,dt$$

$$=\int_{-R}^R\frac{\cosh pt\cosh i\pi p-\sinh pt \sinh i\pi p}{\cosh t}\,dt=\cosh i\pi pI_1=\cos \pi p I_1$$

  • $I_2$, the integral up the right side of the rectangle, is $$\int_0^\pi f(R+it)i\,dt=O(\exp(-(1-|p|)R))\to 0$$

  • $I_4$, the integral down the left hand side of the rectangle, is $$\int_0^\pi -f(-R+it)i\,dt=O(\exp(-(1-|p|)R))\to 0$$

Finally, the integral around the contour is just $2\pi i\text{Res}(f,i\pi/2)=2\pi i \frac{\cosh(i\pi p/2)}{\sinh i\pi/2}=2\pi\cos\frac{\pi p}{2}$.

Noting that as $R\to\infty,I_1\to2I(p):$

$$(1+\cos\pi p)(2I(p))=2\pi\cos\frac{\pi p}{2}\implies I(p)=\frac{\pi}{2\cos\frac{\pi p}{2}}$$

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    It might be easier use a semicircular "keyhole" contour and stay with the integrand as it is. – 2017-01-02
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    @zhw. for me it's a matter of taste; I prefer to avoid branch cuts / logarithms where possible - I consider straight lines and rectangles easier to work with (this is not intended to sound snarky). I also feel this method makes the $p\to-p$ symmetry a little more transparent. – 2017-01-02