Loading [MathJax]/jax/output/HTML-CSS/jax.js

Saturday, February 12, 2022

Integral of the day XXV

An integral involving the reciprocal beta function

Integral related to the reciprocal beta function


Today we show the proof of this integral posted by @integralsbot π0x(sin(x2)cos(x2))sinxdx=2πln(2) The proof will rely on the integral representation of the reciprocal beta function.

Proof:

Recall the following integral representation of the reciprocal beta function (Proof in the Appendix): π0sinv1xsin(ax)dx=πsin(πa2)2v1vB(1+a+v2,1a+v2)v,aR,v>0 π0sinv1xcos(ax)dx=πcos(πa2)2v1vB(1+a+v2,1a+v2)v,aR,v>0 If we put v=12 π0sin(ax)sinxdx=πsin(πa2)2121vB(1+a2+14,1a2+14) π0cos(ax)sinxdx=πcos(πa2)2121vB(1+a2+14,1a2+14) Differentiating with respect to a: ddaπ0sin(ax)sinxdx=π0xcos(ax)sinxdx=2π2cos(πa2)+2πsin(πa2)ψ(34a2)2πsin(πa2)ψ(a2+34)B(a2+34,34a2) ddaπ0cos(ax)sinxdx=π0xsin(ax)sinxdx=2π2sin(πa2)2πcos(πa2)ψ(34a2)+2πcos(πa2)ψ(a2+34)B(a2+34,34a2) Taking the limit as a12 π0xcos(x2)sinxdx=πγ+π2+πψ(12)2 π0xsin(x2)sinxdx=πγπ2+πψ(12)2 From the formula ψ(12J)=ψ(12+J)=γln(4)+Jj=122j1J=0,1,2,... if we put J=0 we have ψ(12)+γ=ln(4) Therefore π0xcos(x2)sinxdx=π2πln(4)2 π0xsin(x2)sinxdx=π2πln(4)2 Hence π0x(sin(x2)cos(x2))sinxdx=π22+πln(4)2π22+πln(4)2=2πln(2) Therefore we can conclude π0x(sin(x2)cos(x2))sinxdx=2πln(2)

Appendix:

Here is the proof of the following integrals with a,bR: π0sinv1xsin(ax)dx=πsin(πa2)2v1vB(1+a+v2,1a+v2)v>0 π0sinv1xcos(ax)dx=πcos(πa2)2v1vB(1+a+v2,1a+v2)v>0

Proof:

Let a,bR I=π0sinv1xeiaxdx=π0eiax(eix+eix2i)v1dx=1(2i)v1π0eiaxj=0(v1j)(1)jeix(v12j)dx=1(2i)v1j=0(v1j)(1)jπ0eix(a+v12j)dx=1(2i)v1j=0(v1j)(1)ji(a+v12j)iπ(a+v12j)0ewdw=1(2i)v1j=0(v1j)(1)j(eiπ(a+v12j)1)i(a+v12j)=1(2i)v1j=0(v1j)(1)j(eiπ(a+v1)1)i(a+v12j)=(eiπ(a+v1)1)2v1ivj=0(v1j)(1)j(a+v12j)=(eiπ(a+v1)1)2v1ivj=0(vj)jj!(1)j(a+v12j)=(eiπ(a+v1)1)2v1ivj=0(1v)j(a+v12j)j! Note that 2jav+1=(a+v1)(3av2)j(av+12)j Hence I=(eiπ(a+v1)1)2v1ivj=0(1v)j(a+v12j)j!=(eiπ(a+v1)1)2v1iv1(a+v1)j=0(1v)j(av+12)j(3av2)j=(eiπ(a+v1)1)2v1iv1(a+v1)F(1v,av+12;3av2;1) From the Gauss's Hypergeometric Theorem: F(a,b;c;1)=Γ(c)Γ(cab)Γ(ca)Γ(cb)c>a+b Therefore, supposing v>0: I=(eiπ(a+v1)1)2v1iv1(a+v1)Γ(3av2)Γ(v)Γ(1a+v2)=(eiπ(a+v1)1)2viv1(1av)2(1av2)Γ(1av2)Γ(v)Γ(1a+v2)=(eiπ(a+v1)1)2vivΓ(1av2)Γ(v)Γ(1a+v2) Note (eiπ(a+v1)1)iv=2sin(πa2)cos(π(a+v)2) (eiπ(a+v1)1)iv=2cos(πa2)sin(π(a+v)2) and recall the Euler's reflection formula for cos(πt): Γ(12+t)Γ(12t)=πcos(πt) Therefore π0sinv1xsin(ax)dx=(π0sinv1xeiaxdx)=sin(πa2)cos(π(a+v)2)2v1Γ(1av2)Γ(v)Γ(1a+v2)=sin(πa2)2v1πΓ(1av2)Γ(v)Γ(1+a+v2)Γ(1a+v2)Γ(1a+v2)(Euler's reflection formula)=sin(πa2)2v1πΓ(v)Γ(1+a+v2)Γ(1a+v2)=πsin(πa2)2v1vB(1+a+v2,1a+v2) In a very similar way we can show : π0sinv1xcos(ax)dx=(π0sinv1xeiaxdx)=πcos(πa2)2v1vB(1+a+v2,1a+v2)

No comments:

Post a Comment

Series of the day

Series involving the digamma and the zeta functions The sum 1(n+1)pnq ...