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Wednesday, August 11, 2021

Generalized hypergeometric function

Integral found on Twitter

Solution to integral involving 3F2


We show the proof of the following result posted by @SrinivasR1729 0ln(1+ex)1exex2dx=2πln(12+12)


Proof

Recall that pFq(a1,a2,...,ap;b1,b2,...,bq;x)=j=0(a1)(a2)(ap)(b1)(b2)(bq)xjj!
where the symbol (x)j refers to the Pochhammer polynomial (also called rising factorial): (x)j=x(x+1)(x+2)(x+j1)
Hence, 0ln(1+ex)1exex2dx=10ln(1+w)1wwdw(wew)=10k=1(1)k+1wkk11wwdw=k=1(1)k+1k10wk12(1w)12dw=k=1(1)k+1kB(k+12,12)=j=0(1)jj+1B(j+32,12)(k=j+1)=j=0(1)jj+1Γ(j+32)Γ(12)Γ(j+2)=j=0(1)jj+1Γ(32)Γ(12)(32)jΓ(2)(2)j(Γ(t+j)=Γ(t)(t)j)=π2j=0(1)jj+1(32)j(2)j(Γ(12)=π,Γ(32)=π2)=π2j=0(1)j(32)j(1)j(2)j(2)j((1)j(2)j=11+j)=π2j=0(1)j(32)j(1)j(1)j(2)j(2)j(1)j=π2j=0(32)j(1)j(1)j(2)j(2)j(1)jj!(j!=(1)j)=π23F2(32,1,1;2,2;1)=2πln(12+12)
Therefore 0ln(1+ex)1exex2dx=2πln(12+12)

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