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.

  • 2
    $\begingroup$ Good question and well-formatted for a first time user. (+1) $\endgroup$
    – Ron Gordon
    Apr 17, 2013 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, 2013 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, 2013 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, 2013 at 21:03
  • $\begingroup$ @user1923: so I've been duped?!? ;) $\endgroup$
    – Ron Gordon
    Apr 17, 2013 at 21:04

1 Answer 1


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.

  • $\begingroup$ 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, 2013 at 20:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.