∫10√−lnxdx
I'm looking for alternative methods to what I already know (method I have used below) to evaluate this Integral.
y=−lnx
ey=e−lnx=eln1x=1x
dx=−dyey
∫10√−lnxdx=∫0∞√y(−dyey)=∫∞0e−yy12dy=(12)!
(12)!=12√π
∫10√−lnxdx=12√π
Answer
As far as I know, there are essentially two ways for proving that Γ(12)=√π.
The first way is to use integration by parts, leading to Γ(z+1)=zΓ(z), in order to relate Γ(12) to the Wallis product. The latter can be computed by exploiting the Weierstrass product for the sine function:
sinzz=+∞∏n=1(1−z2π2n2)
by evaluating it in z=π2. The duplication formula for the Γ function follows from this approach.
The second way is to use some substitutions in order to relate Γ(12) to the gaussian integral
∫+∞−∞e−x2dx
that can be evaluated through Fubini's theorem and polar coordinates.
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