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Can anyone show a proof of $$\int_0^\infty t^{a-1}e^{it}\,dt=\Gamma(a)e^{ia\pi/2}$$ where $0<a<1$, and $$\Gamma(a)=\int_0^\infty t^{a-1}e^{-t}\,dt.$$ Thank you.

JohnD
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TCL
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1 Answers1

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Consider an integral in the complex plane along the contour which goes from the origin out along the real axis to $+R$, then moves in a circular arc up to the point $+iR$, and then returns to the origin along the imaginary axis.

Jonathan
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  • Or go up to $\frac {(R,R)} {\sqrt 2}$ and back to the origin. Shouldn't that work @Jonathan? – superAnnoyingUser Feb 16 '13 at 21:09
  • That contour would not work with the integrand I have in mind (which is the natural one for this problem), but perhaps with a different integrand. – Jonathan Feb 16 '13 at 21:14