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Few days ago, I saw this post by Vivek Kaushik and I began thinking about other possible approach based on it.

We know that: $$\int_{0}^{+\infty}y\cdot e^{-y^2}dy = \frac 1 2\qquad\text{and}\qquad \int_{0}^{+\infty}\frac{1}{1+x^2}dx=\frac \pi 2$$ Follows: $$\int_{0}^{+\infty}\int_{0}^{+\infty}\frac{y\cdot e^{-y^2}}{1+x^2}dydx=\frac \pi 4$$ Clearly, this is the square of the Gaussian Integral: $$\int_{0}^{+\infty} e^{-y^2}dy=\frac{\sqrt{\pi}}{2}$$ So, I started wondering if, with some substitution we can get to something of the form: $$\left(\int_{0}^{+\infty}e^{-y^2}dx\right)^2$$ But, are we able to express that $\frac{y}{1+x^2}$ as the result of another integral involving $e^{-z^2}$? The problem, here, is that I can write: $$\int_{0}^{+\infty}e^{-z\cdot f(x, y)}dz=\frac{y}{1+x^2}$$ where $f(x, y)=\frac{y}{1+x^2}$, but then we would have: $$\int_{0}^{+\infty}\int_{0}^{+\infty}\int_{0}^{+\infty}e^{-y^2}\cdot e^{-z\cdot f(x, y)}\,dzdydx$$ which I can't really simplify.

Thanks.

Theo Bendit
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Matteo
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1 Answers1

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Apply the variable changes $x=\frac t{\sqrt{1+s^2}}$ and $y=\frac {ts}{\sqrt{1+s^2}}$, along with the corresponding Jacobian $\frac t{1+s^2}$

$$\int_{0}^{\infty}e^{-x^2}dx\int_{0}^{\infty}e^{-y^2}dy = \int_{0}^{\infty} \int_{0}^{\infty} e^{-x^2-y^2}dxdy= \int_{0}^{\infty} \int_{0}^{\infty} \frac{te^{-t^2}}{1+s^2}dtds $$

Matteo
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Quanto
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