Gaussian Integral Derivation

Question

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:

1. $F'(x) + G'(x) = 0 $
2. $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!

Answer 1

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?

Answer 2

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...

Answer 3

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, $$ \begin{aligned} G^\prime(x) &=\int_0^1\partial_x\frac{e^{-(1+t^2)x^2}}{1+t^2}\,\mathrm dt\\ &=-\int_0^1(\partial_x x^2)e^{-(1+t^2)x^2}\,\mathrm dt\\ &=-2e^{-x^2}\int_0^1e^{-(tx)^2}x\,\mathrm dt\\ &=-2e^{-x^2}I(x). \end{aligned} $$