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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=∫_0 ^( ∞) ((sin((√( x ))))/( (( e^x ))^(1/4) ))dx=?
$$ \\ $$$$\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{sin}\left(\sqrt{\:{x}\:}\right)}{\:\sqrt[{\mathrm{4}}]{\:{e}^{{x}} }}{dx}=? \\ $$$$ \\ $$
Answered by MrGaster last updated on 31/Mar/25
Let t=(√x)⇒x=t^2 ,dx=2t dt ∫_0 ^∞ sin(t)e^(−t^2 /4) ∙2t dt u=sin(t),du=2te^(−t^2 /4) dt du=cos(t)dt,v=−4e^(−t^2 /4) −4e^(−t^2 /4 ) sin(t)dt,v=−4e^(−t^2 /4) −4e^(−t^2 /4) sin(t)∣_0 ^∞ +4∫_0 ^∞ e^(−t^2 /4) cos(t)dt =4∫_0 ^∞ e^(−t^2 /4) cos(t)dt ∫_0 ^∞ e^(−t^2 ) cos(bt)dt=(1/2)(√(π/a))e^(−b^2 /(4a)) a=(1/4),b=1⇒∫_0 ^∞ e^(−t^2 /4) cos(t)dt=(1/2) =(1/2)∙2(√π)∙e^(−1) =(√π)∙(1/e) 4∙(√π)∙(1/e)=((4(√π))/e)
$$\mathrm{Let}\:{t}=\sqrt{{x}}\Rightarrow{x}={t}^{\mathrm{2}} ,{dx}=\mathrm{2}{t}\:{dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \mathrm{sin}\left({t}\right){e}^{−{t}^{\mathrm{2}} /\mathrm{4}} \centerdot\mathrm{2}{t}\:{dt} \\ $$$${u}=\mathrm{sin}\left({t}\right),{du}=\mathrm{2}{te}^{−{t}^{\mathrm{2}} /\mathrm{4}} {dt} \\ $$$${du}=\mathrm{cos}\left({t}\right){dt},{v}=−\mathrm{4}{e}^{−{t}^{\mathrm{2}} /\mathrm{4}} \\ $$$$−\mathrm{4}{e}^{−{t}^{\mathrm{2}} /\mathrm{4}\:} \mathrm{sin}\left({t}\right){dt},{v}=−\mathrm{4}{e}^{−{t}^{\mathrm{2}} /\mathrm{4}} \\ $$$$−\mathrm{4}{e}^{−{t}^{\mathrm{2}} /\mathrm{4}} \mathrm{sin}\left({t}\right)\mid_{\mathrm{0}} ^{\infty} +\mathrm{4}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} /\mathrm{4}} \mathrm{cos}\left({t}\right){dt} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} /\mathrm{4}} \mathrm{cos}\left({t}\right){dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } \mathrm{cos}\left({bt}\right){dt}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{{a}}}{e}^{−{b}^{\mathrm{2}} /\left(\mathrm{4}{a}\right)} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{4}},{b}=\mathrm{1}\Rightarrow\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} /\mathrm{4}} \mathrm{cos}\left({t}\right){d}\mathrm{t}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{2}\sqrt{\pi}\centerdot{e}^{−\mathrm{1}} =\sqrt{\pi}\centerdot\frac{\mathrm{1}}{{e}} \\ $$$$\mathrm{4}\centerdot\sqrt{\pi}\centerdot\frac{\mathrm{1}}{{e}}=\frac{\mathrm{4}\sqrt{\pi}}{{e}} \\ $$
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)
$$\:\:\:\:\:\mathrm{I}=\:\mathscr{L}\:\left\{\:{sin}\left(\sqrt{{x}}\:\right)\right\}\mid_{{s}=\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\sqrt{\pi}}{\mathrm{2}{s}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{e}^{−\frac{\mathrm{1}}{\mathrm{4}{s}}} \mid_{{s}=\frac{\mathrm{1}}{\mathrm{4}}} =\:\frac{\sqrt{\pi}}{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }\:{e}^{−\mathrm{1}} \\ $$$$\:\:\:\:\:=\:\mathrm{4}\:\frac{\sqrt{\pi}}{{e}} \\ $$

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