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I have to show that $$ \int_{0}^{\infty}\frac{x^{p-1}}{1+x}dx = \frac{\pi}{\sin(p\pi)}; 0<p<1$$

The hint provided says that I should choose as my contour the region between two circles of radius $\epsilon$ and $R$ and integrate $f(\mathbb{z}) = \frac{\mathbb{z}^{p-1}}{1+\mathbb{z}}$

Why such a contour? Shouldn't I be trying to choose some sort of semi-circles to eliminate the integral as $R \rightarrow \infty$ ?

I wanted to understand HOW to decide what contour to use in this problem and others. Once that is decided, I can proceed. Is there a good strategy to arrive at the right contour? Or is it just something you get better at the more problems you see?

Sebastiano
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s_a94248
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1 Answers1

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I gues, you have to use "Key-Hole contour" with branch cut on positive real axes. The next you are solving integrals and two of them will be your initial by simplifying. Two integral, which countours was seleceted on "circles $\epsilon$ and $R$" will disapper (proving by inequality).
Finally you have to use Cauchy Residue Theorem, in your function $f(z)$ you have simple pole $z=-1=e^{\pi i}$
For more information check this video: https://www.youtube.com/watch?v=9H3NS0D5gng
P.S.
The semi circle countour is just one of popular contour because Jordan's Lemma says to us, that integrals kind of $\lim\limits_{R\to \infty}\oint_{C_R} f(z) e^{iaz} \,dz=0$, where $a>0$ and $f(z)$ is continuous in this area formed by the contour.

s_a94248
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Nifton
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  • Do you mean Cauchy instead of Koshi? (=: – PinkyWay Aug 12 '23 at 08:57
  • Yes, thank you. – Nifton Aug 12 '23 at 09:15
  • Please don't answer duplicate questions. – José Carlos Santos Aug 12 '23 at 09:15
  • I didn't know, sorry. But i think that question is new because his main question is in the last paragraph to which I answered to some extent in P.S. – Nifton Aug 12 '23 at 09:20
  • @Nifton I wanted to understand how to decide what contours to use. Is there a good strategy to arrive at the right contour? Or is it just something you get better at the more problems you see? – s_a94248 Aug 12 '23 at 09:52
  • @s_a94248 It really comes with experience. However, it is necessary to look at which function is in the integral, for example, a complex logarithm or a polynomial. It should also be noted that, generally speaking, the initial function should not coincide with the function of the complex variable in structure; this is done on purpose. It is always necessary to monitor the analyticity of a function and, if necessary, use known theorems. That is, it is necessary to choose not only the "correct" circuit, but the "correct" function of the complex variable. – Nifton Aug 12 '23 at 15:30
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    @Nifton right. Yeah, I suppose it will become clearer with time. I am new to complex analysis. But it is fascinating how these integrals can be trivialized in the C-plane compared to doing them only on the real axis. And to think all this is possible just because someone thought to give a value to $\sqrt{-1}$ Amazing! – s_a94248 Aug 12 '23 at 17:19