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Question Number 53695 by gunawan last updated on 25/Jan/19

If ∫_( 0) ^∞  e^(−x^2 ) dx = (√(π/2)) , then ∫_( 0) ^∞  e^(−ax^2 ) dx,  a > 0  is

$$\mathrm{If}\:\underset{\:\mathrm{0}} {\overset{\infty} {\int}}\:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\:\sqrt{\frac{\pi}{\mathrm{2}}}\:,\:\mathrm{then}\:\underset{\:\mathrm{0}} {\overset{\infty} {\int}}\:{e}^{−{ax}^{\mathrm{2}} } {dx}, \\ $$ $${a}\:>\:\mathrm{0}\:\:\mathrm{is} \\ $$

Commented byAbdo msup. last updated on 25/Jan/19

first ∫_0 ^∞  e^(−x^2 ) dx =((√π)/2) (so there is a error at the Q.)  and ∫_0 ^∞  e^(−ax^2 ) dx =_((√a)x=t)    ∫_0 ^∞   e^(−t^2 ) (dt/(√a))  =(1/(√a)) ∫_0 ^∞   e^(−t^2 ) dt =(1/(√a)) ((√π)/2) =((√π)/(2(√a)))    .

$${first}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:\left({so}\:{there}\:{is}\:{a}\:{error}\:{at}\:{the}\:{Q}.\right) \\ $$ $${and}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ax}^{\mathrm{2}} } {dx}\:=_{\sqrt{{a}}{x}={t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}^{\mathrm{2}} } \frac{{dt}}{\sqrt{{a}}} \\ $$ $$=\frac{\mathrm{1}}{\sqrt{{a}}}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}^{\mathrm{2}} } {dt}\:=\frac{\mathrm{1}}{\sqrt{{a}}}\:\frac{\sqrt{\pi}}{\mathrm{2}}\:=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{{a}}}\:\:\:\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19

t=ax^2   x=((√t)/(√a))  dx=(1/(2(√a)))×(1/(√t))dt  ∫_0 ^∞ e^(−t) ×(dt/(2(√a)))×(1/(√t))  (1/(2(√a)))∫_0 ^∞ e^(−t) ×t^((1/2)−1) dt  gamma function ∫_0 ^∞ e^(−t) t^(n−1) dt=⌈(n)  ⌈(n+1)=n!  [factorial of n]  ⌈((1/2))=(√π)   so  (1/(2(√a)))∫_0 ^∞ e^(−t) ×t^((1/2)−1) dt  =(1/(2(√a) ))×(√π)   let calculate ∫_0 ^∞ e^(−x^2 ) dx  y=x^2    x=(√y)       dx=(1/(2(√y)))dy  ∫_0 ^∞ e^(−y) ×(1/2)×y^(−(1/2)) dy  (1/2)∫_0 ^∞ e^(−y) y^((1/2)−1) dy  (1/2)×⌈((1/2))=((√π)/2)  so i think ∫_0 ^∞ e^(−x^2 ) dx=((√π)/2) (  not ((√π)/(√2)))

$${t}={ax}^{\mathrm{2}} \:\:{x}=\frac{\sqrt{{t}}}{\sqrt{{a}}}\:\:{dx}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}}}×\frac{\mathrm{1}}{\sqrt{{t}}}{dt} \\ $$ $$\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} ×\frac{{dt}}{\mathrm{2}\sqrt{{a}}}×\frac{\mathrm{1}}{\sqrt{{t}}} \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} ×{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dt} \\ $$ $${gamma}\:{function}\:\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} {t}^{{n}−\mathrm{1}} {dt}=\lceil\left({n}\right) \\ $$ $$\lceil\left({n}+\mathrm{1}\right)={n}!\:\:\left[{factorial}\:{of}\:{n}\right] \\ $$ $$\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}\: \\ $$ $${so} \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} ×{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dt} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}}\:}×\sqrt{\pi}\: \\ $$ $${let}\:{calculate}\:\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$ $${y}={x}^{\mathrm{2}} \:\:\:{x}=\sqrt{{y}}\:\:\:\:\:\:\:{dx}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{y}}}{dy} \\ $$ $$\int_{\mathrm{0}} ^{\infty} {e}^{−{y}} ×\frac{\mathrm{1}}{\mathrm{2}}×{y}^{−\frac{\mathrm{1}}{\mathrm{2}}} {dy} \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{y}} {y}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dy} \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}}×\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$ $${so}\:{i}\:{think}\:\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}}\:\left(\:\:\boldsymbol{{not}}\:\frac{\sqrt{\pi}}{\sqrt{\mathrm{2}}}\right) \\ $$

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