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Question Number 154409 by mnjuly1970 last updated on 18/Sep/21

    nice calculus..       prove  that :      I:=∫_0 ^( ∞) (( (1+e^( −x)  )sin^( 2) (x))/x^( (3/2)) ) =(√(2π)) ( 1+ (√((√2) − 1)) )   m.n

$$ \\ $$$$\:\:{nice}\:{calculus}.. \\ $$$$\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$\: \\ $$$$\:\mathrm{I}:=\int_{\mathrm{0}} ^{\:\infty} \frac{\:\left(\mathrm{1}+{e}^{\:−{x}} \:\right){sin}^{\:\mathrm{2}} \left({x}\right)}{{x}^{\:\frac{\mathrm{3}}{\mathrm{2}}} }\:=\sqrt{\mathrm{2}\pi}\:\left(\:\mathrm{1}+\:\sqrt{\sqrt{\mathrm{2}}\:−\:\mathrm{1}}\:\right) \\ $$$$\:{m}.{n} \\ $$

Answered by Kamel last updated on 18/Sep/21

   I:=∫_0 ^( ∞) (( (1+e^( −x)  )sin^( 2) (x))/x^( (3/2)) )       =2∫_0 ^1 ∫_0 ^(+∞) x^(−(1/2)) (1+e^(−x) )sin(2ax)dxda     =2∫_0 ^1 ((√2)a^(−(1/2)) ∫_0 ^(+∞) sin(u^2 )du+(1/(2i))∫_0 ^(+∞) (e^(−(1−2ai)x^2 ) −e^(−(1+2ai)x^2 ) )dx)da     =2∫_0 ^1 (2^(−(1/2)) a^(−(1/2)) ((√π)/(2(√2)))+((√π)/(4i))((1/( (√(1−2ai))))−(1/( (√(1+2ai))))))da    =(√π)((1/2)((√(1−2i))+(√(1+2i))))=(√(2π(1+(√5))))

$$ \\ $$$$\:\mathrm{I}:=\int_{\mathrm{0}} ^{\:\infty} \frac{\:\left(\mathrm{1}+{e}^{\:−{x}} \:\right){sin}^{\:\mathrm{2}} \left({x}\right)}{{x}^{\:\frac{\mathrm{3}}{\mathrm{2}}} }\: \\ $$$$\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{+\infty} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+{e}^{−{x}} \right){sin}\left(\mathrm{2}{ax}\right){dxda} \\ $$$$\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sqrt{\mathrm{2}}{a}^{−\frac{\mathrm{1}}{\mathrm{2}}} \int_{\mathrm{0}} ^{+\infty} {sin}\left({u}^{\mathrm{2}} \right){du}+\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{+\infty} \left({e}^{−\left(\mathrm{1}−\mathrm{2}{ai}\right){x}^{\mathrm{2}} } −{e}^{−\left(\mathrm{1}+\mathrm{2}{ai}\right){x}^{\mathrm{2}} } \right){dx}\right){da} \\ $$$$\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}}} {a}^{−\frac{\mathrm{1}}{\mathrm{2}}} \frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\sqrt{\pi}}{\mathrm{4}{i}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{2}{ai}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{2}{ai}}}\right)\right){da} \\ $$$$\:\:=\sqrt{\pi}\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{1}−\mathrm{2}{i}}+\sqrt{\mathrm{1}+\mathrm{2}{i}}\right)\right)=\sqrt{\mathrm{2}\pi\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)} \\ $$

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