I have heard that: $$\int_0^1\frac{x^a-x^{-a}}{x-1}dx=\frac1 a-\pi\cot(\pi a)$$ when $-1<a<1$. How would I prove this?

That doesn't have an elementary indefinite integral, but the definite integral is quite simple.

Someone suggested I use complex analysis to prove it, but I am relatively new to the subject. (I have gotten up to contour integrals, but I'm not sure how to use them to evaluate this particular integral.) I also tried expanding it with a series, but it didn't help.

  • $\begingroup$ Do you know the digamma function $\psi$? $\endgroup$ – Olivier Oloa Jul 25 '14 at 23:18
  • $\begingroup$ I know it's definition and it's recurrence relation. How would it help here? $\endgroup$ – Akiva Weinberger Jul 25 '14 at 23:19
  • $\begingroup$ It has an anti derivative in terms of the hypergeometric function ${}_2 F_1$ $\endgroup$ – ClassicStyle Jul 25 '14 at 23:28
  • $\begingroup$ Idea: Let $\frac{p}{q}=a$ for some $p,q$ integers. Then, there should be some substitution that turns our integral into a rational function. It might be easier from there. (This only works when $a$ is rational, but the irrationals should follow from continuity or some such.) $\endgroup$ – Akiva Weinberger Jul 26 '14 at 0:05

I also tried expanding it with a series, but it didn't help.

It should have helped. If we write

$$\frac{x^{-a}-x^a}{1-x} = (x^{-a}-x^a)\sum_{n=0}^\infty x^n,$$

by the monotone convergence theorem, we have

$$\begin{align} \int_0^1 \frac{x^{-a}-x^a}{1-x}\,dx &= \sum_{n=0}^\infty \int_0^1 (x^{-a}-x^a)x^n\,dx\\ &= \sum_{n=0}^\infty \left(\frac{1}{n+1-a} - \frac{1}{n+1+a}\right)\\ &= - \sum_{n=1}^\infty \left(\frac{1}{a-n}+\frac{1}{a+n}\right)\\ &= \frac{1}{a} - \pi \cot (\pi a) \end{align}$$

by the well-known partial fraction decomposition of the cotangent.

  • 1
    $\begingroup$ That partial fraction decomposition is new to me. I'll add it to my "mathematical toolbox," and i'll see if I can find a simple proof online. $\endgroup$ – Akiva Weinberger Jul 26 '14 at 0:07
  • $\begingroup$ Maybe this could be fleshed out into a simple proof. All you need (for real arguments) is that $\pi\cot (\pi x) - \frac{1}{x}$ can be continuously extended to $0$ by the value $0$, that the series is periodic with period $1$, and that both, the cotangent and the series satisfy the functional equation. $\endgroup$ – Daniel Fischer Jul 26 '14 at 0:13

It could be a way to use the digamma function $\psi$ which verify $$ \psi(a+1)=-\gamma+\int_0^1 \frac{1-x^a}{1-x} \mathrm{d}x, \quad |a|<1,$$ since $$\psi(a)-\psi(1-a)=-\pi \cot (\pi a),$$ $$\psi(1+a)-\psi(a)=\frac{1}{a}.$$

But this is not elementary.

I doubt that you have to prove all from the beginning.

  • 1
    $\begingroup$ @user166353 The gamma function's reflection formula $\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$ induces a similar formula relating $\psi(z)$ and $\psi(1-z)$. $\endgroup$ – Ian Jul 25 '14 at 23:28
  • $\begingroup$ Can you prove the reflection formula you have there? $\endgroup$ – Akiva Weinberger Jul 25 '14 at 23:40
  • $\begingroup$ @user166353 It depends on what you admit. If you know that $\Gamma(z)\Gamma(1-z)=...$ you take logarithms and differentiate. $\endgroup$ – Olivier Oloa Jul 25 '14 at 23:44
  • $\begingroup$ I don't know the proof for that either. (Well, I know the one with the Weierstrass product, but how to prove that that is equivalent to the Gamma function?) I doubt that this is the only way to show this; there must be a simpler way. $\endgroup$ – Akiva Weinberger Jul 25 '14 at 23:46

HINT: Denote the integral by $ I $. Since $$ x^a-x^{-a}=e^{a\ln(x)}-e^{-a\ln(x)}=2\sinh(a\ln x) .$$

Substituting $\ln x= u $, we get $$ I=\int_{-\infty}^0 \frac{ 2\sinh au}{e^u-1 }e^udu $$

To change the limits of the integral to more convenient one, again substitue $ u=-t$ we have \begin{align} I&=\int_0^{\infty}\frac {2\sinh at}{e^{-t }-1} e^{-t} dt \\ &=\int_0^{\infty}\frac{2\sinh at}{e^{t/2}(e^{-t/2}-e^{t/2})}dt \\ &=\int_0^{ \infty }e^{-t/2}\frac {\sinh at}{\sinh (t/2)} dz \end{align} Now consider the function $$ f (z)=\frac{e^{(a-\frac{1}{2})z}}{\sinh (z/2)} $$ and integrate over the closed rectangle $C: \pm R, i R$. Don't forget to jump by a small contours on $0, i\pi $. Clearly, $ f (z) $ is analytic on and inside the contour $ C $. Hence by Cauchy Goursat Thm. $\int_{c } f (z) dz=0$. . .


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