Solution to integral involving 3F2
We show the proof of the following result posted by @SrinivasR1729 ∫∞0ln(1+e−x)√1−e−xe−x2dx=2πln(12+1√2)
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+j−1)
Hence,
∫∞0ln(1+e−x)√1−e−xe−x2dx=∫10ln(1+w)√1−w√wdw(w↦e−w)=∫10∞∑k=1(−1)k+1wkk1√1−w√wdw=∞∑k=1(−1)k+1k∫10wk−12(1−w)−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)=π2∞∑j=0(−1)jj+1(32)j(2)j(Γ(12)=√π,Γ(32)=√π2)=π2∞∑j=0(−1)j(32)j(1)j(2)j(2)j((1)j(2)j=11+j)=π2∞∑j=0(−1)j(32)j(1)j(1)j(2)j(2)j(1)j=π2∞∑j=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+1√2)
Therefore
∫∞0ln(1+e−x)√1−e−xe−x2dx=2πln(12+1√2)
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