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

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

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2019-05-18 22:05:11
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Here is an illustration of just how we can review the indispensable without complex analysis.

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

Split the array right into 2 periods $(0,1)$ and also $(1,\infty)$ after that make use of the replacement $x=1/t$ in the last component to get

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

Expand the integrand as a power collection making use of $(1-x)^{-1} = \sum_{n=0}^\infty x^n$ and also incorporate to get

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

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

we keep in mind 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 also the outcome adheres to given that the indispensable concerned is $I(a)-I(b).$

0
2019-05-21 21:21:15
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I assume it could be like the evidence of $B(x,1-x)=\pi\csc(\pi x)$. Allow $x=\exp(y)$ after that review the indispensable $\int_{-\infty}^\infty \frac{\exp(ay)-\exp(by)}{\exp(y)-1}=\pi(\cot(a\pi)-\cot(b\pi))$ making use of shape assimilation.

I'll leave the information as a workout (that I can not be troubled finishing!)

0
2019-05-21 07:44:47
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To round off Simon is solution, see Example 4.3.3 (p. 244-- 245) in Complex Variables: Introduction and Applications by Ablowitz & Fokas.

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2019-05-21 07:37:23
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This looks evocative Frullani is Integrals ; see as an example the article:

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2019-05-21 07:22:49
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