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I need help in integrating this integral:

$$\int_{-\infty}^{\infty} e^{-2b x^2} \, dx$$

Knowing the value of $\int_{-\infty}^{\infty} e^{-x^2} \, dx$, and the integral for $e^{bx}$, I'm guessing that it would go out and evaluate into something like:

$$\int_{-\infty}^{\infty} e^{-2b x^2} \, dx= \frac{\sqrt\pi}{-2b}$$

But Wolfram gives the integral out as:

$$\int_{-\infty}^{\infty} e^{-2b x^2} \, dx= \frac{\sqrt\frac{\pi}{2}}{\sqrt{b}}, Re[b] \gt 0$$

Which has $b$ under the radical sign and a factor of $\sqrt{\frac{1}{2}}$ on the numerator.

mopy
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2 Answers2

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Assuming that $b\in\mathbb{R}$ and $b>0$, you would use the substitution $w=\sqrt{2b}\,x$, so that $2bx^2=w^2$. Since $dw=\sqrt{2b}\,dx$, we find $$ \int_{-\infty}^{\infty}e^{-2bx^2}\,dx=\frac{1}{\sqrt{2b}}\int_{-\infty}^{\infty}e^{-w^2}\,dw=\frac{1}{\sqrt{2b}}\sqrt{\pi}=\sqrt{\frac{\pi}{2b}}. $$

Nick Peterson
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if you view the integral as a function of $b \gt 0$, say $$ I(b)=\int_{-\infty}^{\infty} e^{-2b x^2} \, dx $$

then integrating by parts gives $$ I = 4b\int_{-\infty}^{\infty}x^2e^{-2bx^2}\,dx $$ on the other hand, differentiating $I(b)$ wrt $b$ we find $$ \frac{dI}{db}=-2\int_{-\infty}^{\infty}x^2e^{-2bx^2}\,dx $$ putting these together: $$ \frac{dI}{db} = -\frac12 \frac{I}{b} $$ indefinite integration gives $$ I(b) = \frac{k}{\sqrt{b}} $$ for some constant $k$. setting $b=\frac14$ we have $$ 2k = \int_{-\infty}^{\infty} e^{-\frac{x^2}2} \, dx = \sqrt{2\pi} $$ so, finally $$ I(b) = \sqrt{\frac{\pi}{2b}} $$

David Holden
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