Find: ∫_0 ^( ∞) sin^2 ( (√x) ) e^(−x) dx = ?


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Question Number 197360 by hardmath last updated on 14/Sep/23

$Find :$ $\int_{0}^{\infty} \sin^{2} (\sqrt{x}) e^{- x} dx = ?$
	Answered by Mathspace last updated on 15/Sep/23
	$I=∫_0 ^∞ ((1−cos(2(√x)))/2) e^(−x) dx =(1/2)∫_0 ^∞ e^(−x) dx−(1/2)∫_0 ^∞ cos(2(√x))e^(−x) dx ∫_0 ^∞ e^(−x) dx=[−e^(−x) ]_0 ^∞ =1 ∫_0 ^∞ cos(2(√x))e^(−x) dx ((√x)=t) =∫_0 ^∞ cos(2t)e^(−t^2 ) (2t)dt =[−e^(−x^2 ) cos(2t)]_0 ^∞ −2∫_0 ^∞ e^(−t^2 ) sin(2t)dt =1−2∫_0 ^∞ e^(−t^2 ) sin(2t)dt but ∫_0 ^∞ e^(−t^2 ) sin(2t)dt =Im(∫_0 ^∞ e^(−t^2 +2it) dt) and ∫_0 ^∞ e^(−t^2 +2it) dt=∫_0 ^∞ e^(−(t^2 −2it +i^2 −i^2 )) dt =∫_0 ^∞ e^(−(t−i)^2 +1) dt (t−i=y) =e ∫_(−i) ^∞ e^(−y^2 ) dy=eλ_0 erf(−i)⇒ ∫_0 ^∞ sin^2 ((√x))e^(−x) dx =(1/2)−(1/2){1−2 Im(eλ_o erf(−i)} =Im(eλ_0 erf(−i)} ....be continued...$
	$I = \int_{0}^{\infty} \frac{1 - c o s (2 \sqrt{x})}{2} e^{- x} d x$ $= \frac{1}{2} \int_{0}^{\infty} e^{- x} d x - \frac{1}{2} \int_{0}^{\infty} c o s (2 \sqrt{x}) e^{- x} d x$ $\int_{0}^{\infty} e^{- x} d x = {[- e^{- x}]}_{0}^{\infty} = 1$ $\int_{0}^{\infty} c o s (2 \sqrt{x}) e^{- x} d x (\sqrt{x} = t)$ $= \int_{0}^{\infty} c o s (2 t) e^{- t^{2}} (2 t) d t$ $= {[- e^{- x^{2}} c o s (2 t)]}_{0}^{\infty} - 2 \int_{0}^{\infty} e^{- t^{2}} s i n (2 t) d t$ $= 1 - 2 \int_{0}^{\infty} e^{- t^{2}} s i n (2 t) d t$ $b u t \int_{0}^{\infty} e^{- t^{2}} s i n (2 t) d t$ $= I m (\int_{0}^{\infty} e^{- t^{2} + 2 i t} d t)$ $a n d \int_{0}^{\infty} e^{- t^{2} + 2 i t} d t = \int_{0}^{\infty} e^{- (t^{2} - 2 i t + i^{2} - i^{2})} d t$ $= \int_{0}^{\infty} e^{- {(t - i)}^{2} + 1} d t (t - i = y)$ $= e \int_{- i}^{\infty} e^{- y^{2}} d y = e λ_{0} e r f (- i) \Rightarrow$ $\int_{0}^{\infty} {s i n}^{2} (\sqrt{x}) e^{- x} d x$ $= \frac{1}{2} - \frac{1}{2} {1 - 2 I m (e λ_{o} e r f (- i)}$ $= I m (e λ_{0} e r f (- i)}$ $. . . . b e c o n t i n u e d . . .$
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