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Wednesday, June 16, 2021

Feynman's trick

Feynman

Integral involving the Feynman's integral trick

Today we show the proof of this result which involves the famous Feynman's integral trick twice. The post of @infseriesbot has a typo (the sign is wrong). First, we differentiate the parameter a under the integral sign: dda0(1cosax)ln(x)x2dx=0dda(1cosax)ln(x)x2dx=0sinaxln(x)xdx=0sinwln(wa)wdw(wax)=0sinwln(w)wdwln(a)0sinwwdw=0sinwln(w)wdwπln(a)2 =0sinwddtwt|t=0+wdwπln(a)2=ddt|t=0+0wt1sinwdwπln(a)2 Here we are using the fact that ddtwt|t=0+=ln(w) and the fact that 0sin(w)wdw=π2 (a famous result obtained through complex analysis). Note that the first integral is the Mellin transform of sinw, so we could use the following expansion series of sin (a consequence of the Euler's formula and the de Moivre theorem) in oder to apply the Ramanujan's master theorem: sinw=n=0(w)n[sin(nπ2)]n! Then, by Ramanujan's master theorem: ddt|t=0+0wt1sinwdw=ddt|t=0+[Γ(t)sin(tπ2)]=limt0+Γ(t)[π2cos(tπ2)+ψ0(t)sin(πt2)]=limt0+Γ(t+1)[π2cos(tπ2)t+ψ0(t)sin(πt2)t] Update: I found a cool way to calculate this limit:
Note that π2cos(tπ2)t=π21tπ323t2!+π525t34!π727t56!+...=π21t+O(t) Also note that ψ0(t)sin(πt2)t=ψ0(t)π2ψ0(t)π323t23!+ψ0(t)π525t45!ψ0(t)π727t67!+...=ψ0(t)π2tψ0(t)(π323t3!π525t35!+π727t57!...)=ψ0(t)π2tψ0(t)O(t) Therefore π2cos(tπ2)t+ψ0(t)sin(πt2)t=π21t+ψ0(t)π2+O(t)tψ0(t)O(t)=π2(1t+ψ0(t))+O(t)tψ0(t)O(t)=π2ψ0(t+1)+O(t)tψ0(t)O(t) limt0+Γ(t+1)[π2cos(tπ2)t+ψ0(t)sin(πt2)t]=limt0+Γ(t+1)[π2ψ0(t+1)+O(t)tψ0(t)O(t)]=π2γ
where limt0+ψ0(t+1)=γ and limt0+tψ0(t)=1

Therefore
dda0(1cosax)ln(x)x2dx=π2γπln(a)20(1cosax)ln(x)x2dx=π2aγ+aln(a)da+C=aπγ2aπln(a)2+πa2+C Note that if a=0 then C=0 therefore 0(1cosax)ln(x)x2dx=πa2(γ+ln(a)1)

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