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How to integrate $\int_{-\infty}^\infty e^{\large \frac{-x^2}{a^2}} dx$ using substitution?

I know that $\int_0^\infty e^{-x^2}dx = \frac12\sqrt{\pi}$.


marked as duplicate by tired, Morgan Rodgers, Kamil Jarosz, Tom-Tom, Harish Chandra Rajpoot Feb 1 '16 at 0:01

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  • 4
    $\begingroup$ Try substituting $ t = x / a $ $\endgroup$ – Jytug Jan 31 '16 at 14:26
  • $\begingroup$ Duplicate Alert!!!!!!!!!!!!! $\endgroup$ – tired Jan 31 '16 at 17:29
  • $\begingroup$ @tired Actually, the answers there don't explain the above at all: they deal with a more general question, and this particular step is treated as obvious. $\endgroup$ – user147263 Jan 31 '16 at 18:06

Notice, $$\int_{-\infty}^{\infty}e^{-x^2/a^2}\ dx$$ using property of even function, $f(-x)=f(x)$, $$=2\int_{0}^{\infty}e^{-x^2/a^2}\ dx$$ Now, let $\frac{x}{a}=t\implies \frac{dx}{a}=dt$ or $dx=a\ dt$, $$=2\int_{0}^{\infty}e^{-t^2} (a\ dt)$$ $$=2a\int_{0}^{\infty}e^{-t^2} dt$$ we know $\int_{0}^{\infty}e^{-x^2} dx=\frac{\sqrt \pi}{2}$, $$=2a\cdot \frac{\sqrt \pi}{2}$$ $$=\color{red}{a\sqrt \pi}$$


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