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I'm asked to show the following equality given $a\in (-1,1)\subset\Bbb R$

$$\int\limits_0^\infty\frac{x^a\ \log(x)}{(1+x)^2}dx=\frac{\pi\sin(\pi a)-a\pi^2\cos(\pi a)}{\sin^2(a\pi)}$$

So I'm trying to use the keyhole contour (as shown here) and so far I've been able to see the integrals along the circunferences tend to $0$ as $R\to\infty$ and $\varepsilon\to 0$. Computing $\text{Res}[f(z);-1]$ (where $f(z)$ equals the integrand) I get

$$\lim_{z\to -1}\frac{d}{dz}(1+z)^2f(z)=\lim_{z\to -1}az^{a-1}\log(z)+z^{a-1}=e^{\pi i(a-1)}(a(1+\pi i)+1)$$

On the other side,

$$\begin{align*}\int_\varepsilon^Rf(z)dz+\int_R^\varepsilon f(z)dz&=\int_\varepsilon^R\frac{z^a\log|z|}{(1+z)^2}dz+\int_R^\varepsilon\frac{z^a(\log|z|+2\pi i)}{(1+z)^2}dz\\ &=\int_\varepsilon^R\frac{z^a\log|z|}{(1+z)^2}dz+\int_R^\varepsilon\frac{z^a\log|z|}{(1+z)^2}dz-2\pi i\int_\varepsilon^R\frac{z^a}{(1+z)^2}\end{align*}$$

I know the result of the last integral, but I'm not sure whether what I've done is right, and what to do to finish it. Any help is appreciated.

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    $\begingroup$ Good question and well-formatted for a first time user. (+1) $\endgroup$ – Ron Gordon Apr 17 '13 at 19:00
  • $\begingroup$ In the answer of Ron Gordon you can find "I leave to the reader to show that $\int_0^{\infty} dx \frac{x^a}{(1+x)^2} = \frac{\pi a}{\sin{\pi a}}$". Differentiating both sides by $a$ and changing the order of integration and derivation you obtain the required equality. I leave the details for the reader :-) $\endgroup$ – vesszabo Apr 17 '13 at 20:44
  • $\begingroup$ @RonGordon Actually, to be fair, I'm not a first time user; I've been around for a while. $\endgroup$ – user1923 Apr 17 '13 at 20:55
  • $\begingroup$ @vesszabo: true, but nice to know that you get the correct result from actually doing out the contour integral. $\endgroup$ – Ron Gordon Apr 17 '13 at 21:03
  • $\begingroup$ @user1923: so I've been duped?!? ;) $\endgroup$ – Ron Gordon Apr 17 '13 at 21:04
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The point of the keyhole contour is to exploit the multivaluedness of the function being integrated. In this case, the function is $z^a \log{z}$. The right-hand side should look like

$$\int_0^{\infty} dx \frac{x^a \log{x}}{(1+x)^2} - e^{i 2 \pi a} \int_0^{\infty} dx \frac{x^a (\log{x}+i 2 \pi)}{(1+x)^2}$$

So you were halfway there - you had the $i 2 \pi$ from the log, but you also needed the other factor from the $z^a$.

The integral over the circular arcs vanish in the limits of $R \rightarrow \infty$ and $\epsilon \rightarrow 0$. It appears you are attempting to show this. I will leave the rest of the details to the reader.

Now, I leave it to the reader to show that

$$\int_0^{\infty} dx \frac{x^a}{(1+x)^2} = \frac{\pi a}{\sin{\pi a}}$$

(which result you claim to have), so that

$$\left (1-e^{i 2 \pi a}\right) \int_0^{\infty} dx \frac{x^a \log{x}}{(1+x)^2} - i 2 \pi e^{i 2 \pi a} \frac{\pi a}{\sin{\pi a}} = i 2 \pi (-1) e^{i \pi a} (1+i \pi a)$$

(I get $1+i \pi a$, not $1+(1+i \pi)a$, in the residue.)

Do out the algebra - the sought result follows.

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  • $\begingroup$ For some weird reason I wrote $\log(1)=1$ in the residue, and forgot the $e^{i2\pi a}$. Anyway, thanks for your help. $\endgroup$ – user1923 Apr 17 '13 at 20:55

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