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Question Number 28756 by abdo imad last updated on 29/Jan/18

find in terms of λ ∫_0 ^∞ e^(−λt) ((sint)/(√t)) dt with λ>0

$${find}\:{in}\:{terms}\:{of}\:\lambda\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\lambda{t}} \:\frac{{sint}}{\sqrt{{t}}}\:{dt}\:\:{with}\:\:\lambda>\mathrm{0} \\ $$

Commented byabdo imad last updated on 30/Jan/18

let put I= ∫_0 ^∞ e^(−λt) ((sint)/(√t))dt with λ>0 the ch.(√t) =x give I= ∫_0 ^∞ e^(−λx^2 ) ((sin(x^2 ))/x) 2xdx =2∫_0 ^∞ e^(−λx^2 ) sin(x^2 )dx = ∫_(−∞) ^(+∞) e^(−λx^2 ) sin(x^2 )dx=− Im (∫_(−∞) ^(+∞) e^(−λx^2 −ix^2 ) dx) but ∫_(−∞) ^(+∞) e^(−λx^2 −ix^2 ) dx = ∫_(−∞) ^(+∞) e^(−(λ+i)x^2 ) dx= ∫_(−∞) ^(+∞) e^(−((√(λ+i))x)^2 ) dx = ∫_(−∞ ) ^(+∞) e^(−u^2 ) (du/(√(λ+i))) (ch.(√(λ+i)) x=u) =(λ +i)^(−(1/2)) .(√π) we have λ+i= (√(1+λ^2 )) ( (λ/(√(1+λ^2 ))) +(i/(√(1+λ^2 ))) ) =r e^(iθ) ⇒r=(√(1+λ^2 )) and cosθ= (λ/(√(1+λ^2 ))) and sinθ= (1/((√(1+λ^2 )) )) ⇒tanθ=(1/λ) ⇒θ= arctan((1/λ)) so λ+i=(1+λ^2 )^(1/2) e^(iarctan((1/λ))) λ+i= (1+λ^2 )^(1/2) e^(i((π/2) −arctanλ)) ⇒ (λ+i)^(−(1/2)) =(1+λ^2 )^(−(1/4)) e^(i( ((−π)/4) +((arctanλ)/2))) =(1+λ^2 )^(−(1/4)) (cos( ((−π)/4) +((arctanλ)/2)) +isin(((−π)/4)+((arctanλ)/2))) I=−Im((√π)(λ+i)^(−(1/2)) )=−(√π)sin(((−π)/4) +((artanλ)/2)) =(√π) sin((π/4) −((arctanλ)/2)) .

$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\lambda{t}} \:\:\frac{{sint}}{\sqrt{{t}}}{dt}\:\:\:{with}\:\:\lambda>\mathrm{0}\:{the}\:{ch}.\sqrt{{t}}\:={x}\:{give} \\ $$ $${I}=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\lambda{x}^{\mathrm{2}} \:\:} \frac{{sin}\left({x}^{\mathrm{2}} \right)}{{x}}\:\mathrm{2}{xdx}\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:{e}^{−\lambda{x}^{\mathrm{2}} } \:{sin}\left({x}^{\mathrm{2}} \right){dx} \\ $$ $$=\:\int_{−\infty} ^{+\infty} \:{e}^{−\lambda{x}^{\mathrm{2}} } \:{sin}\left({x}^{\mathrm{2}} \right){dx}=−\:{Im}\:\left(\int_{−\infty} ^{+\infty} \:\:{e}^{−\lambda{x}^{\mathrm{2}} \:−{ix}^{\mathrm{2}} } {dx}\right)\:{but} \\ $$ $$\int_{−\infty} ^{+\infty} \:\:{e}^{−\lambda{x}^{\mathrm{2}} \:−{ix}^{\mathrm{2}} } {dx}\:=\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left(\lambda+{i}\right){x}^{\mathrm{2}} } {dx}=\:\int_{−\infty} ^{+\infty} \:{e}^{−\left(\sqrt{\lambda+{i}}{x}\right)^{\mathrm{2}} } \:{dx}\:\:\: \\ $$ $$=\:\int_{−\infty\:} ^{+\infty} \:\:{e}^{−{u}^{\mathrm{2}} } \:\:\frac{{du}}{\sqrt{\lambda+{i}}}\:\:\:\:\:\:\:\:\left({ch}.\sqrt{\lambda+{i}}\:{x}={u}\right) \\ $$ $$=\left(\lambda\:+{i}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} .\sqrt{\pi}\:\:\:\:\:\:\:\:\:{we}\:{have} \\ $$ $$\lambda+{i}=\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\left(\:\:\frac{\lambda}{\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} \:}}\:\:+\frac{{i}}{\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}\:\right)\:={r}\:{e}^{{i}\theta} \:\Rightarrow{r}=\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} } \\ $$ $${and}\:{cos}\theta=\:\frac{\lambda}{\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} \:}}\:\:\:\:{and}\:{sin}\theta=\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:}\:\:\Rightarrow{tan}\theta=\frac{\mathrm{1}}{\lambda} \\ $$ $$\Rightarrow\theta=\:{arctan}\left(\frac{\mathrm{1}}{\lambda}\right)\:{so}\:\lambda+{i}=\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{{iarctan}\left(\frac{\mathrm{1}}{\lambda}\right)} \\ $$ $$\lambda+{i}=\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{{i}\left(\frac{\pi}{\mathrm{2}}\:−{arctan}\lambda\right)} \\ $$ $$\Rightarrow\:\left(\lambda+{i}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{{i}\left(\:\frac{−\pi}{\mathrm{4}}\:+\frac{{arctan}\lambda}{\mathrm{2}}\right)} \\ $$ $$=\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:\left({cos}\left(\:\frac{−\pi}{\mathrm{4}}\:+\frac{{arctan}\lambda}{\mathrm{2}}\right)\:+{isin}\left(\frac{−\pi}{\mathrm{4}}+\frac{{arctan}\lambda}{\mathrm{2}}\right)\right) \\ $$ $${I}=−{Im}\left(\sqrt{\pi}\left(\lambda+{i}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)=−\sqrt{\pi}{sin}\left(\frac{−\pi}{\mathrm{4}}\:+\frac{{artan}\lambda}{\mathrm{2}}\right) \\ $$ $$=\sqrt{\pi}\:{sin}\left(\frac{\pi}{\mathrm{4}}\:−\frac{{arctan}\lambda}{\mathrm{2}}\right)\:. \\ $$

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