Gaussian Integral Derivation I am currently working on the derivation of the Gaussian Integral $\int_0^\infty e^{-t^2} dt$.
For this I have two functions F, G: $[0, \infty) \mapsto \mathbb{R} $ with:
$$F(X) = (\int_0^x e^{-t^2} dt )^2 $$ and $$G(X) = \int_0^1 \frac{e^{-x^2(1 + t^2)}}{1 + t^2} dt$$
Now I am supposed to show the following two things:

*

*$F'(x) + G'(x) = 0 $

*$F(x) + G(x) = \pi/4 $
for all x in $[0, \infty)$.

I am completely lost with this exercises and unfortunately, don't have any further hints.
Can you please help me in solving these two exercises? Thank you!
 A: Hints:

*

*For 1: To $F(x)$, apply the chain rule (outer function: squaring, inner function: an accumulation integral) and the fundamental theorem of calculus.

*For 1: To $G(x)$ apply the Leibniz rule since you want to differentiate under the integral sign.

*For 2: In part 1, you showed $F+G$ is constant.  Is there any value of $x$ at which you can evaluate both integrals?  ... say a choice of $x$ where the $F$ integral is trivial and the $G$ doesn't have an exponential any more...

A: HINT:
$$\begin{align}
G'(x)&=\frac{d}{dx}\int_0^1 \frac{e^{-x^2(1+t^2)}}{1+t^2}\,dt\\\\
&=\frac{d(x^2)}{dx}\frac{d}{d(x^2)}\int_0^1 \frac{e^{-x^2(1+t^2)}}{1+t^2}\,dt\\\\
&=-2x\int_0^1 e^{-x^2(1+t^2)}\,dt\\\\
\end{align}$$
Can you finish now by making a substitution of a variable?
A: Hint:
For brevity let $I(x)=\int_0^xe^{-t^2}\,\mathrm dt$. We have, by Leibniz's rule and the chain rule
$$
F^\prime(x)=2I(x)I^\prime(x)=2e^{-x^2}I(x).
$$
Furthermore,
$$
\begin{aligned}
G^\prime(x)
&=\int_0^1\partial_x\frac{e^{-(1+t^2)x^2}}{1+t^2}\,\mathrm dt\\
&=-\int_0^1(\partial_x x^2)e^{-(1+t^2)x^2}\,\mathrm dt\\
&=-2e^{-x^2}\int_0^1e^{-(tx)^2}x\,\mathrm dt\\
&=-2e^{-x^2}I(x).
\end{aligned}
$$
