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Compute

$$\int_0^\infty \frac{x}{(1+x^2)x^\alpha}dx, \space \alpha \in (0,1).$$

So I thought of writing it as

$$\int_0^\infty \frac{x^{1-\alpha}}{1+x^2}dx$$ But since $\alpha \in (0,1)$ a simple $u$-sub wouldn't work. Unless I let $u = 1+x^2$ then $du = 2xdx$ thus $dx = \frac{du}{2x}$ and I get

$$\int_1^\infty \frac{x^{1-\alpha}}{u2x}du=\int_1^\infty \frac{1}{u2x^\alpha}du$$

Then I get stuck, unless I use by parts ?

MyMathYourMath
  • 4,591

1 Answers1

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Usually integrals of this type are handled using residue calculus from complex analysis. A more general result from Exercise 7, Page 208 of Gamelin's Complex Analysis is that $$\int_0^{\infty}\frac{x^{a-1}}{1+x^b}dx=\frac{\pi}{b\sin(\pi a/b)},\enspace 0<a<b.$$ Then it can be deduced $$\int_0^{\infty}\frac{x^{1-a}}{1+x^2}=\frac{\pi}{2\sin\big(\pi(2-a)/2\big)}=\frac{\pi}{2\sin\big(\pi a/2\big)}.$$

Brian
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