In THIS ANSWER, I used straightforward contour integration to evaluate the integral ∫∞0xa1+x2dx=π2sec(πa2)for |a|<1.
An alternative approach is to enforce the substitution x→ex to obtain
∫∞0xa1+x2dx=∫∞−∞e(a+1)x1+e2xdx=∫0−∞e(a+1)x1+e2xdx+∫∞0e(a−1)x1+e−2xdx=∞∑n=0(−1)n(∫0−∞e(2n+1+a)xdx+∫∞0e−(2n+1−a)xdx)=∞∑n=0(−1)n(12n+1+a+12n+1−a)=2∞∑n=0(−1)n(2n+1(2n+1)2−a2)=π2sec(πa2)
Other possible ways forward include writing the integral of interest as
∫∞0xa1+x2dx=∫10xa+x−a1+x2dx
and proceeding similarly, using 11+x2=∑∞n=0(−1)nx2n.
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 |a|<1. Letting x=tanθ, we have
∫∞0xa1+x2dx=∫π/20tanaθdθ=∫π/20sinaθcos−aθdθ
The last integral is half the beta integral B((a+1)/2,(1−a)/2), Thus
∫π/20sinaθcos−aθdθ=12Γ(a+12)Γ(1−a2)Γ(a+12+1−a2)=12Γ(a+12)Γ(1−a2)
By Euler reflection,
Γ(a+12)Γ(1−a2)=πcsc[π(1+a2)]=πsec(πa2)
and the result follows.
Edit: For a proof of Euler reflection without contour integration, start with the integral function f(x)=∫∞0ux−1(1+u)−1du, and show that f solves the differential equation y″y−(y′)2=y4, y(1/2)=π, y′(1/2)=0. The solution is πcscπx. On the other hand, f(x) is the beta integral B(1+x,1−x), which is equal to Γ(x)Γ(1−x). I believe this method is due to Dedekind.
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