∫_0 ^( π^2 ) x^(−(1/2)) e^(−x) dx


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Question Number 127211 by mohammad17 last updated on 27/Dec/20

$\int_{0}^{π^{2}} x^{- \frac{1}{2}} e^{- x} d x$
	Answered by Dwaipayan Shikari last updated on 27/Dec/20

	$\int_{0}^{π^{2}} x^{- \frac{1}{2}} e^{- x} d x x = u^{2} \Rightarrow 1 = 2 u \frac{d u}{d x}$ $= 2 \int_{0}^{π} e^{- u^{2}} d u = 2 \int_{0}^{\infty} e^{- u^{2}} d u - 2 \int_{π}^{\infty} e^{- u^{2}} d u$ $= \sqrt{π} - \frac{2 e^{- π^{2}}}{2 π + \frac{1}{π + \frac{2}{2 π + \frac{3}{π + . . .}}}}$ $O r \int_{0}^{π^{2}} x^{- \frac{1}{2}} e^{- x} d x = \int_{0}^{\infty} x^{- \frac{1}{2}} e^{- x} - \int_{π^{2}}^{\infty} x^{- \frac{1}{2}} e^{- x^{2}} = Γ (\frac{1}{2}) - Γ (\frac{1}{2}, π^{2})$ $I n c o m p l e t e G a m m a Γ (a, z) = \int_{z}^{\infty} t^{a - 1} e^{- t} d t$
	Commented by mohammad17 last updated on 27/Dec/20

	$s i r i s t h e v a l u e i n f i n t e o r f i n i t e ?$
	Commented by Dwaipayan Shikari last updated on 27/Dec/20
	$Finite, It is a continued fraction Q127186$
	$F i n i t e, I t i s a c o n t i n u e d f r a c t i o n$ $Q 127186$
	Answered by mindispower last updated on 28/Dec/20

	$\sum_{k ⩾ 0} \int^{π^{2}} x^{- \frac{1}{2}} . \frac{{(- x)}^{k}}{k!} d x$ $= \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{k!} \int_{0}^{π^{2}} x^{k - \frac{1}{2}} d x$ $= \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{k!} {[\frac{x^{k + \frac{1}{2}}}{k + \frac{1}{2}}]}_{0}^{π} = \sum_{k ⩾ 0} \frac{{(- 1)}^{k} π^{2 k + 1}}{(k + \frac{1}{2}) k!}$ $= 2 π + 2 π \sum_{k ⩾ 1} \frac{\underset{j = 0}{\prod^{k - 1}} (\frac{1}{2} + j)}{\underset{j = 0}{\prod^{k - 1}} (\frac{3}{2} + j)} . \frac{{(- π^{2})}^{k}}{k!}$ $= 2 π (1 + \sum_{k ⩾ 1} \frac{{(\frac{1}{2})}_{k}}{{(\frac{3}{2})}_{k}} . \frac{{(- π^{2})}^{k}}{k!})$ $= 2 π_{1} F_{1} (\frac{1}{2}; \frac{3}{2}; - π^{2})$
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