# about the gaussian integral $\int_0^{\infty} e^{-x^2}\ \mathrm{d}x$

I have an exercise which seem to be a method of calculating the Gaussian integral :

Let $\displaystyle f(x)=\int_0^1 \frac{\exp(-x^2(t^2+1))}{t^2+1}\ \mathrm{d}t$

Study the deriviability of $f$, then conclude that $\displaystyle \int_0^{\infty} e^{-x^2}\ \mathrm{d}x=\frac{\sqrt{\pi}}{2}$.

I'm stuck in the second question : by Leibinz rule $f$ is differentiable over $\mathbb{R}$, but I couldn't calculate $f'(x)$ in terms of $x$ only or figure out the relation between the two questions.

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Are you sure the upper integration limit is 1? –  Ron Gordon Jan 23 '13 at 15:24
gowers.wordpress.com/2007/10/04/… –  user58512 Jan 23 '13 at 15:44

Consider $F(t)=\left(\displaystyle \int_0^t e^{-x^2}\ \mathrm{d}x\right)^2$.
Differentiate with respect to $t$ and set $x=ty$. You will arrive at:$$F{\prime}(t)=-{\frac{d}{dt}}\int_0^1 {\frac{e^{-(1+y^2)t^2}}{1+y^2}}\ \mathrm{d}y$$
Write this as: $F^{\prime}(t)=-G^{\prime}(t)$ so there is a constant $C$ such that $F(t)=-G(t)+C$ for all $t>0$.
To find $C$ let $t$ to tent to $0$. The left hand side goes to 0 obviously while the right hand side goes to ${\pi}/4+C$. Hence $C=-{\pi}/4$. Substitute in $F(t)=-G(t)+C$ and let t to tend to infinity to arrive at what you need.