Question Number 197360 by hardmath last updated on 14/Sep/23
$$\mathrm{Find}: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{sin}^{\mathrm{2}} \:\left(\:\sqrt{\mathrm{x}}\:\right)\:\mathrm{e}^{−\boldsymbol{\mathrm{x}}} \:\mathrm{dx}\:=\:? \\ $$
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...](https://www.tinkutara.com/question/Q197366.png)
$${I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{cos}\left(\mathrm{2}\sqrt{{x}}\right)}{\mathrm{2}}\:{e}^{−{x}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {cos}\left(\mathrm{2}\sqrt{{x}}\right){e}^{−{x}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {dx}=\left[−{e}^{−{x}} \right]_{\mathrm{0}} ^{\infty} =\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\infty} {cos}\left(\mathrm{2}\sqrt{{x}}\right){e}^{−{x}} {dx}\:\:\:\:\left(\sqrt{{x}}={t}\right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} {cos}\left(\mathrm{2}{t}\right){e}^{−{t}^{\mathrm{2}} } \left(\mathrm{2}{t}\right){dt} \\ $$$$=\left[−{e}^{−{x}^{\mathrm{2}} } {cos}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{\infty} −\mathrm{2}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } \:{sin}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{1}−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{2}} } {sin}\left(\mathrm{2}{t}\right){dt} \\ $$$${but}\:\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } {sin}\left(\mathrm{2}{t}\right){dt} \\ $$$$={Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{2}} +\mathrm{2}{it}} {dt}\right) \\ $$$${and}\:\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} +\mathrm{2}{it}} {dt}=\int_{\mathrm{0}} ^{\infty} {e}^{−\left({t}^{\mathrm{2}} −\mathrm{2}{it}\:+{i}^{\mathrm{2}} −{i}^{\mathrm{2}} \right)} \:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({t}−{i}\right)^{\mathrm{2}} +\mathrm{1}} {dt}\:\:\:\left({t}−{i}={y}\right) \\ $$$$={e}\:\int_{−{i}} ^{\infty} \:{e}^{−{y}^{\mathrm{2}} } {dy}={e}\lambda_{\mathrm{0}} {erf}\left(−{i}\right)\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} {sin}^{\mathrm{2}} \left(\sqrt{{x}}\right){e}^{−{x}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}−\mathrm{2}\:{Im}\left({e}\lambda_{{o}} {erf}\left(−{i}\right)\right\}\right. \\ $$$$={Im}\left({e}\lambda_{\mathrm{0}} {erf}\left(−{i}\right)\right\} \\ $$$$….{be}\:{continued}… \\ $$