# Evaluating the improper integral $\int\limits_{0}^{\infty} \frac{x^{a-1} - x^{b-1}}{1-x} \ dx$

How does one evaluate the integral $$\int\limits_{0}^{\infty} \frac{x^{a-1} - x^{b-1}}{1-x} \ dx \quad \text{for} \ a,b \in (0,1)$$

Here's a sketch of how we can evaluate the integral without complex analysis.

$$\text{Let} \quad I(a) = \int_0^\infty \frac{x^{a-1}}{1-x} dx.$$

Split the range into two intervals $(0,1)$ and $(1,\infty)$ then use the substitution $x=1/t$ in the latter part to obtain

$$I(a) = \int_0^1 \frac{x^{a-1}-x^{-a}}{1-x} dx.$$

Expand the integrand as a power series using $(1-x)^{-1} = \sum_{n=0}^\infty x^n$ and integrate to obtain

$$I(a) = \frac{1}{a} + \sum_{n=1}^\infty \left( \frac{1}{a+n} + \frac{1}{a-n} \right).$$

Now, by differentiating logarithmically the product formula for $\sin x,$ $$\sin \pi x = \pi x \prod_{n=1}^{\infty} \left( 1 - \frac{x^2}{n^2} \right),$$

we note that

$$\pi \cot \pi x = \frac{1}{x} + \sum_{n=1}^\infty \left( \frac{1}{x+n} + \frac{1}{x-n} \right).$$

Thus $$I(a) = \pi \cot(\pi a)$$ and the result follows since the integral in question is $I(a)-I(b).$

• The initial equation given for I(a) requires a pole prescription to be well defined. The second equation implicitly assumes that the Cauchy principle value was used in the definition of I(a). After that it's clear sailing... – Simon Nov 20 '10 at 10:53
• Funny how the cotangent identity keeps popping up lately. – Raskolnikov Feb 22 '11 at 14:45

I think it might be like the proof of $B(x,1-x)=\pi\csc(\pi x)$. Let $x=\exp(y)$ then evaluate the integral $\int_{-\infty}^\infty \frac{\exp(ay)-\exp(by)}{\exp(y)-1}=\pi(\cot(a\pi)-\cot(b\pi))$ using contour integration.

I'll leave the details as an exercise (that I can't be bothered completing!)

This looks reminiscent of Frullani's Integrals; see for example the article:

• Reminiscent, but not quite matching (having $1-x$ in the denominator is quite different from having just $x$). Nice suggestion, still. And by the way: one of the authors, Victor Moll, has written a whole book with this kind of stuff, Irrestistible Integrals (with George Boros). books.google.com/books?id=g9tH8M3NvDIC – Hans Lundmark Oct 17 '10 at 18:54
• Even though it's not a Frullani integral thanks for your post, as when I saw the question I thought to myself, that looks a bit like a f...ll... integral, but the full name just would not come to mind (that's what you get for turning 50). So you saved me some minutes of extra frustration. Thanks! – Derek Jennings Oct 17 '10 at 19:17
• @Hans -- thanks for the great pointer to the book; @Derek: you are welcome! – user1709 Oct 17 '10 at 20:04

To finish off Simon's answer, see Example 4.3.3 (p. 244–245) in Complex Variables: Introduction and Applications by Ablowitz & Fokas.