I=∫_0 ^( ∞) ((sin((√( x ))))/( (( e^x ))^(1/4) ))dx=?


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Question Number 218148 by mnjuly1970 last updated on 30/Mar/25

$I = \int_{0}^{\infty} \frac{\sin (\sqrt{x})}{\sqrt[4]{e^{x}}} d x = ?$
	Answered by MrGaster last updated on 31/Mar/25

	$Let t = \sqrt{x} \Rightarrow x = t^{2}, d x = 2 t d t$ $\int_{0}^{\infty} \sin (t) e^{- t^{2} / 4} \cdot 2 t d t$ $u = \sin (t), d u = 2 {t e}^{- t^{2} / 4} d t$ $d u = \cos (t) d t, v = - 4 e^{- t^{2} / 4}$ $- 4 e^{- t^{2} / 4} \sin (t) d t, v = - 4 e^{- t^{2} / 4}$ $- 4 e^{- t^{2} / 4} \sin (t) ∣_{0}^{\infty} + 4 \int_{0}^{\infty} e^{- t^{2} / 4} \cos (t) d t$ $= 4 \int_{0}^{\infty} e^{- t^{2} / 4} \cos (t) d t$ $\int_{0}^{\infty} e^{- t^{2}} \cos (b t) d t = \frac{1}{2} \sqrt{\frac{π}{a}} e^{- b^{2} / (4 a)}$ $a = \frac{1}{4}, b = 1 \Rightarrow \int_{0}^{\infty} e^{- t^{2} / 4} \cos (t) d t = \frac{1}{2}$ $= \frac{1}{2} \cdot 2 \sqrt{π} \cdot e^{- 1} = \sqrt{π} \cdot \frac{1}{e}$ $4 \cdot \sqrt{π} \cdot \frac{1}{e} = \frac{4 \sqrt{π}}{e}$
	Commented by mnjuly1970 last updated on 31/Mar/25


	Answered by mnjuly1970 last updated on 31/Mar/25
	$I= L { sin((√x) )}∣_(s=(1/4)) = ((√π)/(2s^(3/2) )) e^(−(1/(4s))) ∣_(s=(1/4)) = ((√π)/(2((1/2))^3 )) e^(−1) = 4 ((√π)/e)$
	$I = L {s i n (\sqrt{x})} ∣_{s = \frac{1}{4}}$ $= \frac{\sqrt{π}}{2 s^{\frac{3}{2}}} e^{- \frac{1}{4 s}} ∣_{s = \frac{1}{4}} = \frac{\sqrt{π}}{2 {(\frac{1}{2})}^{3}} e^{- 1}$ $= 4 \frac{\sqrt{π}}{e}$
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