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:
dda∫∞0(1−cosax)ln(x)x2dx=∫∞0dda(1−cosax)ln(x)x2dx=∫∞0sinaxln(x)xdx=∫∞0sinwln(wa)wdw(w↦ax)=∫∞0sinwln(w)wdw−ln(a)∫∞0sinwwdw=∫∞0sinwln(w)wdw−πln(a)2 =∫∞0sinwddtwt|t=0+wdw−πln(a)2=ddt|t=0+∫∞0wt−1sinwdw−πln(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+∫∞0wt−1sinwdw=ddt|t=0+[Γ(t)sin(tπ2)]=limt→0+Γ(t)[π2cos(tπ2)+ψ0(t)sin(πt2)]=limt→0+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=π21t−π323t2!+π525t34!−π727t56!+...=π21t+O(t)
Also note that
ψ0(t)sin(πt2)t=ψ0(t)π2−ψ0(t)π323t23!+ψ0(t)π525t45!−ψ0(t)π727t67!+...=ψ0(t)π2−tψ0(t)(π323t3!−π525t35!+π727t57!−...)=ψ0(t)π2−tψ0(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)
limt→0+Γ(t+1)[π2cos(tπ2)t+ψ0(t)sin(πt2)t]=limt→0+Γ(t+1)[π2ψ0(t+1)+O(t)−tψ0(t)O(t)]=−π2γ
where
limt→0+ψ0(t+1)=−γ and
limt→0+tψ0(t)=−1
Therefore
dda∫∞0(1−cosax)ln(x)x2dx=−π2γ−πln(a)2⟹∫∞0(1−cosax)ln(x)x2dx=−π2∫aγ+aln(a)da+C=−aπγ2−aπln(a)2+πa2+C
Note that if
a=0 then
C=0 therefore
∫∞0(1−cosax)ln(x)x2dx=−πa2(γ+ln(a)−1)
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