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Question Number 97489 by ali_golmakani last updated on 08/Jun/20

please prove it ∫_0 ^∞ e^(−ax^2 ) cos bx dx= (1/2)(√(π/a)).e^(−(b^2 /(4a)))

$${please}\:\:{prove}\:{it} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}^{\mathrm{2}} } \mathrm{cos}\:{bx}\:\:{dx}=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{{a}}}.{e}^{−\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}}} \\ $$

Answered by smridha last updated on 08/Jun/20

=(1/2)Re∫_(−∞) ^∞ e^(−ax^2 +ibx) dx =(1/2)Re(√(𝛑/a)).e^(((−ib)^2 )/(4a)) =(1/2)(√(𝛑/a)).e^(−(b^2 /(4a))) we know ∫_(−∞) ^(+∞) e^(−𝛂x^2 +𝛃x) dx=(√(𝛑/𝛂)).e^(𝛃^2 /(4𝛂)) this is one of the results used in quantum mechanics for finding the probality density of particle. i also prove the result...which i directly used here.

$$=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{R}}{e}\int_{−\infty} ^{\infty} \boldsymbol{{e}}^{−\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{ibx}}} \boldsymbol{{dx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{Re}\sqrt{\frac{\boldsymbol{\pi}}{\boldsymbol{{a}}}}.\boldsymbol{{e}}^{\frac{\left(−\boldsymbol{{ib}}\right)^{\mathrm{2}} }{\mathrm{4}\boldsymbol{{a}}}} \:=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\boldsymbol{\pi}}{{a}}}.{e}^{−\frac{\boldsymbol{{b}}^{\mathrm{2}} }{\mathrm{4}\boldsymbol{{a}}}} \\ $$$$\boldsymbol{{we}}\:\boldsymbol{{know}}\:\int_{−\infty} ^{+\infty} \boldsymbol{{e}}^{−\boldsymbol{\alpha{x}}^{\mathrm{2}} +\boldsymbol{\beta{x}}} \:\boldsymbol{{dx}}=\sqrt{\frac{\boldsymbol{\pi}}{\boldsymbol{\alpha}}}.{e}^{\frac{\boldsymbol{\beta}^{\mathrm{2}} }{\mathrm{4}\boldsymbol{\alpha}}} \\ $$$$\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{one}}\:\boldsymbol{{of}}\:{the}\:\boldsymbol{{results}}\:\boldsymbol{{used}} \\ $$$$\boldsymbol{{in}}\:\boldsymbol{{quantum}}\:\boldsymbol{{mechanics}}\:\boldsymbol{{for}}\:\boldsymbol{{finding}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{probality}}\:\boldsymbol{{density}}\:\boldsymbol{{of}}\:\boldsymbol{{particle}}. \\ $$$$\boldsymbol{{i}}\:\boldsymbol{{also}}\:\boldsymbol{{prove}}\:\boldsymbol{{the}}\:\boldsymbol{{result}}...\boldsymbol{{which}}\: \\ $$$$\boldsymbol{{i}}\:\boldsymbol{{directly}}\:\boldsymbol{{used}}\:\boldsymbol{{here}}. \\ $$$$ \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 08/Jun/20

∫_0 ^∞ e^(−ax^2 ) cos(bx)dx =(1/2)Re(∫_(−∞) ^∞ e^(−ax^2 −ibx) dx) but ∫_(−∞) ^(+∞) e^(−ax^2 −ibx) dx =∫_(−∞) ^(+∞) e^(−a( x^2 +2((ib)/(2a))x +(((ib)/(2a)))^2 −(((ib)/(2a)))^2 )) dx =∫_(−∞) ^(+∞) e^(−a{ (x+((ib)/(2a)))^2 +(b^2 /(4a^2 ))}) dx = e^(−((b2)/(4a))) ∫_(−∞) ^(+∞) e^(−a(x+((ib)/(2a)))^2 ) dx =_(x+((ib)/(2a))=t) e^(−(b^2 /(4a))) ∫_(−∞) ^(+∞) e^(−at^2 ) dt =_((√a)t =u) e^(−(b^2 /(4a))) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/(√a)) =((√π)/(2(√a))) e^(−(b^2 /(4a))) =(1/2)(√(π/a))e^(−(b^2 /(4a))) (a>0)

$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{ax}^{\mathrm{2}} } \mathrm{cos}\left(\mathrm{bx}\right)\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Re}\left(\int_{−\infty} ^{\infty} \:\mathrm{e}^{−\mathrm{ax}^{\mathrm{2}} −\mathrm{ibx}} \:\:\mathrm{dx}\right)\:\mathrm{but} \\ $$$$\int_{−\infty} ^{+\infty} \:\mathrm{e}^{−\mathrm{ax}^{\mathrm{2}} −\mathrm{ibx}} \:\mathrm{dx}\:=\int_{−\infty} ^{+\infty} \:\mathrm{e}^{−\mathrm{a}\left(\:\mathrm{x}^{\mathrm{2}} \:+\mathrm{2}\frac{\mathrm{ib}}{\mathrm{2a}}\mathrm{x}\:\:\:+\left(\frac{\mathrm{ib}}{\mathrm{2a}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{ib}}{\mathrm{2a}}\right)^{\mathrm{2}} \right)} \:\mathrm{dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\mathrm{e}^{−\mathrm{a}\left\{\:\:\left(\mathrm{x}+\frac{\mathrm{ib}}{\mathrm{2a}}\right)^{\mathrm{2}} +\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{4a}^{\mathrm{2}} }\right\}} \:\mathrm{dx}\:\:=\:\mathrm{e}^{−\frac{\mathrm{b2}}{\mathrm{4a}}} \:\int_{−\infty} ^{+\infty} \:\mathrm{e}^{−\mathrm{a}\left(\mathrm{x}+\frac{\mathrm{ib}}{\mathrm{2a}}\right)^{\mathrm{2}} } \mathrm{dx} \\ $$$$=_{\mathrm{x}+\frac{\mathrm{ib}}{\mathrm{2a}}=\mathrm{t}} \:\:\:\:\:\mathrm{e}^{−\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{4a}}} \:\int_{−\infty} ^{+\infty} \:\mathrm{e}^{−\mathrm{at}^{\mathrm{2}} } \mathrm{dt}\:=_{\sqrt{\mathrm{a}}\mathrm{t}\:=\mathrm{u}} \:\:\:\mathrm{e}^{−\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{4a}}} \:\int_{−\infty} ^{+\infty} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \frac{\mathrm{du}}{\sqrt{\mathrm{a}}} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{a}}}\:\mathrm{e}^{−\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{4a}}} \:\:=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{\mathrm{a}}}\mathrm{e}^{−\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{4a}}} \:\:\:\:\:\:\:\left(\mathrm{a}>\mathrm{0}\right) \\ $$

Commented by ali_golmakani last updated on 08/Jun/20

thank you dear friend

$${thank}\:{you}\:{dear}\:{friend} \\ $$

Commented by mathmax by abdo last updated on 08/Jun/20

you are welcome friend

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{friend} \\ $$

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