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How do I show that

$\Gamma(v,x)=(v-1)\Gamma(v-1,x) + x^{v-1} e^{-v}$.

I know I should use Integration by parts on the formula for upper incomplete gamma function. But the result is not coming out. A hint would be helpful, Thanks.

Prince
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1 Answers1

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Did you try the integration by parts? If $x < y$ you have $$\int_x^y t^{v-1}e^{-t} \, dt = - y^{v-1}e^{-y} + x^{v-1} e^{-x} + (v-1) \int_x^y t^{v-2} e^{-t} \, dt.$$ Now let $y \to \infty$ to conclude (since both integrals converge) that $$\int_x^\infty t^{v-1} e^{-t} \, dt = x^{v-1}e^{-x} + (v-1) \int_x^\infty t^{v-2} e^{-t} \, dt.$$

Umberto P.
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  • @Umbero P Thanks very much. My problem was the wrong choice of u and dv in the integration by parts. – Prince Nov 13 '19 at 20:32