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I have a problem to solve and I have something that i don't know how to do.

The half-space Gaussian integral is given :

$$\int_{0}^\infty \exp(-ax^2)dx = \frac{1}2 \sqrt{\frac{\pi}{a}}$$

I have to calculate

$$\int_{0}^\infty \exp \left(-y^2 - \frac{c^2}{y^2} \right)dy$$

So I think we need to use a u-substitution but i can't find something...

Is someone have an idea ? :)

Thanks for your answers

Mathieu

Mathieu Hassenforder
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2 Answers2

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Note that $$I=\int_{0}^{\infty}\exp\left(-y^{2}-\frac{c^{2}}{y^{2}}\right)dy=\frac{1}{2}\int_{-\infty}^{\infty}\exp\left(-y^{2}-\frac{c^{2}}{y^{2}}\right)dy $$ $$=\frac{\exp\left(-2c\right)}{2}\int_{-\infty}^{\infty}\exp\left(-\left(y-\frac{c}{y}\right)^{2}\right)dy $$ and now since $$\int_{-\infty}^{\infty}f\left(x\right)dx=\int_{-\infty}^{\infty}f\left(x-\frac{k}{x}\right)dx,\, k>0 $$ (see here for the proof) we have $$I=\frac{\exp\left(-2c\right)}{2}\int_{-\infty}^{\infty}\exp\left(-y^{2}\right)dy=\color{red}{\frac{\exp\left(-2c\right)\sqrt{\pi}}{2}}.$$

Community
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Marco Cantarini
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Hint: use u-substitution. More explicitly, you need to find a transformation f(y) so that when you substitute u = f(y) into your equation, the integral becomes simpler. The trick with this technique is to notice that your "dy" stays as it is, so you will need to find "du".

The explanation is a little foggy in the abstract, so here's an example: Integrate(y*exp{-y^2} dy), let u=-y^2 which makes du=-2ydy, so my previously difficult integral becomes Integrate((-1/2)exp{u}du) = (-1/2)exp{u} + c. A little back substitution, and the answer is clear. You might try writing this out on paper - then the parallels between your equation and mine should solidify.

Eric
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  • Thanks for your answer :) I already tried with u-substitution by saying : $$u = \frac{-c^2}{y^2}$$ but we obtain an integral too complicate... Do you have an another idea ? – Mathieu Hassenforder Dec 13 '16 at 04:50