# $\int_{0}^{\infty}{dx \over (1+x)x^\alpha} = {\pi \over \sin(\pi (1-\alpha))}$

Can you help me show that

$$\int_{0}^{\infty}{dx \over (1+x)x^\alpha} = {\pi \over \sin(\pi (1-\alpha))}$$

such that $\alpha \lt1$?

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Possible dublicate. – Mhenni Benghorbal Dec 19 '12 at 20:57
Here is a method without contour integration. – Mhenni Benghorbal Dec 19 '12 at 21:00

## 1 Answer

The integral $$\int_{0}^{+\infty}\frac{x^{-\alpha}}{x+1}\, dx$$ converges if $0<\alpha<1$. Take for example $a=-1$. Then, $$\int_{0}^{+\infty}\frac{x}{x+1}\, dx=\int_{0}^{+\infty}1-\frac{1}{x+1}\, dx=\lim_{x\to +\infty}x-\ln(x+1)=+\infty$$ We shall calculate that integral (for $0<\alpha<1$) with complex analysis: The associated complex function is $$f(z)=\frac{z^{-\alpha}}{z+1}$$ where $z^{-\alpha}$ is chosen so that it is continuous and holomorphic in $\mathbb{C}\setminus\left\{(x,0):x\ge 0\right\}$. We shall integrate over the keyhole integral: We have $$\int_{C}f(z)\, dz=\int_{\gamma_ \epsilon}f(z)\, dz+\int_{[\epsilon i,R+\epsilon i]}f(z)\, dz+\int_{\gamma_R}f(z)\, dz+\int_{[ R-\epsilon i,-\epsilon i]}f(z)\, dz$$ The main observation here is that for $x>0$, \begin{gather}\lim_{\epsilon\to 0^+}\log(x+\epsilon i)=\lim_{\epsilon\to 0^+}\log\sqrt{x^2+\epsilon^2}+i\arg(x+\epsilon i)=\log x\notag\\ \lim_{\epsilon\to 0^+}\log(x-\epsilon i)=\lim_{\epsilon\to 0^+}\log\sqrt{x^2+\epsilon^2}+i\arg(x-\epsilon i)=\log x+2\pi i \end{gather} Therefore, $$\int_{[\epsilon i,R+\epsilon i]}f(z)\, dz=\int_{0}^{R}f(x+\epsilon i)\, dx=\int_{0}^{R}\frac{x^{-\alpha}}{x+\epsilon i+1}\, dx= \int_{0}^{R}\frac{e^{\log(x+\epsilon i)(-\alpha)}}{x+\epsilon i+1}\, dx\to\int_{0}^{R}\frac{e^{\log x(-\alpha)}}{x+1}\, dx= \int_{0}^{R}\frac{x^{-\alpha}}{x+1}\, dx$$ as $\epsilon\to 0^+$ while $$\int_{[R-\epsilon i,-\epsilon i]}f(z)\, dz=\int_{R}^{0}f(x-\epsilon i)\, dx=-\int_{0}^{R}\frac{e^{\log(x-\epsilon i)(-\alpha)}}{x-\epsilon i+1}\, dx\to-\int_{0}^{R}\frac{e^{[\log x+2\pi i](-\alpha)}}{x+1}\, dx= -e^{-2\pi i\alpha}\int_{0}^{R}\frac{x^{-\alpha}}{x+1}\, dx$$ again as $\epsilon\to 0^+$.

By the Residue theorem, $$\int_{C}f(z)\, dz=2\pi Res_{-1}(f)=(-1)^{-\alpha}=2\pi e^{-(\log-1)\alpha}=2\pi ie^{-i\pi\alpha}$$ The integrals of $f$ over $\gamma_R$ and $\gamma_{\epsilon}$ converge to $0$ as $R\to +\infty$, $\epsilon\to 0^+$ (why?) and so $$2\pi i e^{-i\pi\alpha}=(1-e^{-2\pi i\alpha})\int_{0}^{+\infty}\frac{x^{-\alpha}}{x+1}\, dx$$ which implies $$\int_{0}^{+\infty}\frac{x^{-\alpha}}{x+1}\, dx=\frac{\pi}{\sin \pi \alpha}={\pi \over \sin(\pi (1-\alpha))}$$

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Thanks! Do you know a method without contour integration? Would perhaps substitution do? – adamG Dec 19 '12 at 17:15
@adamG I very much doubt that any substitution will do. But then again, if there is one, it will be posted here. – Nameless Dec 19 '12 at 17:20
Thanks a lot again! Would you tell me how you did this in 3 minutes:)? – adamG Dec 19 '12 at 17:23
@adamG No magician shares his secrets that easily! Now, this is a classic example of keyhole integration so I already had most of it in my computer. – Nameless Dec 19 '12 at 17:25
I like your answer! Thanks! – adamG Dec 19 '12 at 17:27