Menu Close

Find-e-ax-2-dx-when-a-is-constant-without-changing-the-coordinate-




Question Number 214360 by shunmisaki007 last updated on 06/Dec/24
Find ∫_(−∞) ^∞ e^(−ax^2 ) dx when a is constant without changing the coordinate.
$$\mathrm{Find}\:\underset{−\infty} {\overset{\infty} {\int}}{e}^{−{ax}^{\mathrm{2}} } {dx}\:\mathrm{when}\:{a}\:\mathrm{is}\:\mathrm{constant}\:\mathrm{without}\:\mathrm{changing}\:\mathrm{the}\:\mathrm{coordinate}. \\ $$
Answered by mathmax last updated on 06/Dec/24
=2∫_0 ^∞  e^(−ax^2 ) dx          {a>0}   (√a)x=t  =2∫_0 ^∞  e^(−t^2 ) (dt/( (√a)))=(2/( (√a)))×((√π)/2)=((√π)/( (√a)))
$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ax}^{\mathrm{2}} } {dx}\:\:\:\:\:\:\:\:\:\:\left\{{a}>\mathrm{0}\right\}\:\:\:\sqrt{{a}}{x}={t} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{2}} } \frac{{dt}}{\:\sqrt{{a}}}=\frac{\mathrm{2}}{\:\sqrt{{a}}}×\frac{\sqrt{\pi}}{\mathrm{2}}=\frac{\sqrt{\pi}}{\:\sqrt{{a}}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *