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): ∫π0sinv−1xsin(ax)dx=πsin(πa2)2v−1vB(1+a+v2,1−a+v2)v,a∈R,v>0 ∫π0sinv−1xcos(ax)dx=πcos(πa2)2v−1vB(1+a+v2,1−a+v2)v,a∈R,v>0 If we put v=12 ∫π0sin(ax)√sinxdx=πsin(πa2)212−1vB(1+a2+14,1−a2+14) ∫π0cos(ax)√sinxdx=πcos(πa2)212−1vB(1+a2+14,1−a2+14) Differentiating with respect to a: dda∫π0sin(ax)√sinxdx=∫π0xcos(ax)√sinxdx=√2π2cos(πa2)+√2πsin(πa2)ψ(34−a2)−√2πsin(πa2)ψ(a2+34)B(a2+34,34−a2) dda∫π0cos(ax)√sinxdx=−∫π0xsin(ax)√sinxdx=√2π2sin(πa2)−√2πcos(πa2)ψ(34−a2)+√2πcos(πa2)ψ(a2+34)B(a2+34,34−a2) Taking the limit as a→12 ∫π0xcos(x2)√sinxdx=πγ+π2+πψ(12)2 −∫π0xsin(x2)√sinxdx=πγ−π2+πψ(12)2 From the formula ψ(12−J)=ψ(12+J)=−γ−ln(4)+J∑j=122j−1J=0,1,2,... if we put J=0 we have ψ(12)+γ=−ln(4) Therefore ∫π0xcos(x2)√sinxdx=π2−πln(4)2 −∫π0xsin(x2)√sinxdx=−π2−πln(4)2 Hence ∫π0x(sin(x2)−cos(x2))√sinxdx=π22+πln(4)2−π22+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,b∈R: ∫π0sinv−1xsin(ax)dx=πsin(πa2)2v−1vB(1+a+v2,1−a+v2)v>0 ∫π0sinv−1xcos(ax)dx=πcos(πa2)2v−1vB(1+a+v2,1−a+v2)v>0
Proof:
Let a,b∈R I=∫π0sinv−1xeiaxdx=∫π0eiax(eix+e−ix2i)v−1dx=1(2i)v−1∫π0eiax∞∑j=0(v−1j)(−1)jeix(v−1−2j)dx=1(2i)v−1∞∑j=0(v−1j)(−1)j∫π0eix(a+v−1−2j)dx=1(2i)v−1∞∑j=0(v−1j)(−1)ji(a+v−1−2j)∫iπ(a+v−1−2j)0ewdw=1(2i)v−1∞∑j=0(v−1j)(−1)j(eiπ(a+v−1−2j)−1)i(a+v−1−2j)=1(2i)v−1∞∑j=0(v−1j)(−1)j(eiπ(a+v−1)−1)i(a+v−1−2j)=(eiπ(a+v−1)−1)2v−1iv∞∑j=0(v−1j)(−1)j(a+v−1−2j)=(eiπ(a+v−1)−1)2v−1iv∞∑j=0(v−j)jj!(−1)j(a+v−1−2j)=(eiπ(a+v−1)−1)2v−1iv∞∑j=0(1−v)j(a+v−1−2j)j! Note that 2j−a−v+1=−(a+v−1)(3−a−v2)j(−a−v+12)j Hence I=(eiπ(a+v−1)−1)2v−1iv∞∑j=0(1−v)j(a+v−1−2j)j!=(eiπ(a+v−1)−1)2v−1iv1(a+v−1)∞∑j=0(1−v)j(−a−v+12)j(3−a−v2)j=(eiπ(a+v−1)−1)2v−1iv1(a+v−1)F(1−v,−a−v+12;3−a−v2;1) From the Gauss's Hypergeometric Theorem: F(a,b;c;1)=Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)c>a+b Therefore, supposing v>0: I=(eiπ(a+v−1)−1)2v−1iv1(a+v−1)Γ(3−a−v2)Γ(v)Γ(1−a+v2)=(eiπ(a+v−1)−1)2viv1−(1−a−v)2(1−a−v2)Γ(−1−a−v2)Γ(v)Γ(1−a+v2)=−(eiπ(a+v−1)−1)2vivΓ(1−a−v2)Γ(v)Γ(1−a+v2) Note ℑ(eiπ(a+v−1)−1)i−v=−2sin(πa2)cos(π(a+v)2) ℜ(eiπ(a+v−1)−1)i−v=−2cos(πa2)sin(π(a+v)2) and recall the Euler's reflection formula for cos(πt): Γ(12+t)Γ(12−t)=πcos(πt) Therefore ∫π0sinv−1xsin(ax)dx=ℑ(∫π0sinv−1xeiaxdx)=sin(πa2)cos(π(a+v)2)2v−1Γ(1−a−v2)Γ(v)Γ(1−a+v2)=sin(πa2)2v−1πΓ(1−a−v2)Γ(v)Γ(1+a+v2)Γ(1−a+v2)Γ(1−a+v2)(Euler's reflection formula)=sin(πa2)2v−1πΓ(v)Γ(1+a+v2)Γ(1−a+v2)=πsin(πa2)2v−1vB(1+a+v2,1−a+v2) In a very similar way we can show : ∫π0sinv−1xcos(ax)dx=ℜ(∫π0sinv−1xeiaxdx)=πcos(πa2)2v−1vB(1+a+v2,1−a+v2)
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