Thursday, 26 December 2013

definite integrals - Real-Analysis Methods to Evaluate inti0nftyfracxa1+x2,dx, $|a|




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 xex to obtain




0xa1+x2dx=e(a+1)x1+e2xdx=0e(a+1)x1+e2xdx+0e(a1)x1+e2xdx=n=0(1)n(0e(2n+1+a)xdx+0e(2n+1a)xdx)=n=0(1)n(12n+1+a+12n+1a)=2n=0(1)n(2n+1(2n+1)2a2)=π2sec(πa2)




Other possible ways forward include writing the integral of interest as



0xa1+x2dx=10xa+xa1+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/20sinaθcosaθdθ



The last integral is half the beta integral B((a+1)/2,(1a)/2), Thus



π/20sinaθcosaθdθ=12Γ(a+12)Γ(1a2)Γ(a+12+1a2)=12Γ(a+12)Γ(1a2)



By Euler reflection,



Γ(a+12)Γ(1a2)=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)=0ux1(1+u)1du, and show that f solves the differential equation yy(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,1x), which is equal to Γ(x)Γ(1x). I believe this method is due to Dedekind.


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