In THIS ANSWER, I used straightforward contour integration to evaluate the integral \bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^a}{1+x^2}\,dx=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)}for |a|<1.
An alternative approach is to enforce the substitution x\to e^x to obtain
\begin{align} \int_0^\infty \frac{x^a}{1+x^2}\,dx&=\int_{-\infty}^\infty \frac{e^{(a+1)x}}{1+e^{2x}}\,dx\\\\ &=\int_{-\infty}^0\frac{e^{(a+1)x}}{1+e^{2x}}\,dx+\int_{0}^\infty\frac{e^{(a-1)x}}{1+e^{-2x}}\,dx\\\\ &=\sum_{n=0}^\infty (-1)^n\left(\int_{-\infty}^0 e^{(2n+1+a)x}\,dx+\int_{0}^\infty e^{-(2n+1-a)x}\,dx\right)\\\\ &=\sum_{n=0}^\infty (-1)^n \left(\frac{1}{2n+1+a}+\frac{1}{2n+1-a}\right)\\\\ &=2\sum_{n=0}^\infty (-1)^n\left(\frac{2n+1}{(2n+1)^2-a^2}\right) \tag 1\\\\ &=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)\tag 2 \end{align}
Other possible ways forward include writing the integral of interest as
\begin{align} \int_0^\infty \frac{x^a}{1+x^2}\,dx&=\int_{0}^1 \frac{x^{a}+x^{-a}}{1+x^2}\,dx \end{align}
and proceeding similarly, using \frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}.
Without appealing to complex analysis, what are other approaches one can use to evaluate this very standard integral?
EDIT:
Note that we can show that (1) is the partial fraction representation of (2) using Fourier series analysis. I've included this development for completeness in the appendix of the solution I posted on THIS PAGE.
Answer
I'll assume \lvert a\rvert < 1. Letting x = \tan \theta, we have
\int_0^\infty \frac{x^a}{1 + x^2}\, dx = \int_0^{\pi/2}\tan^a\theta\, d\theta = \int_0^{\pi/2} \sin^a\theta \cos^{-a}\theta\, d\theta
The last integral is half the beta integral B((a + 1)/2, (1 - a)/2), Thus
\int_0^{\pi/2}\sin^a\theta\, \cos^{-a}\theta\, d\theta = \frac{1}{2}\frac{\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right)}{\Gamma\left(\frac{a+1}{2} + \frac{1-a}{2}\right)} = \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right)
By Euler reflection,
\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right) = \pi \csc\left[\pi\left(\frac{1+a}{2}\right)\right] = \pi \sec\left(\frac{\pi a}{2}\right)
and the result follows.
Edit: For a proof of Euler reflection without contour integration, start with the integral function f(x) = \int_0^\infty u^{x-1}(1 + u)^{-1}\, du, and show that f solves the differential equation y''y - (y')^2 = y^4, y(1/2) = \pi, y'(1/2) = 0. The solution is \pi \csc \pi x. On the other hand, f(x) is the beta integral B(1+x,1-x), which is equal to \Gamma(x)\Gamma(1-x). I believe this method is due to Dedekind.
No comments:
Post a Comment