Stuck on crucial step while computing $\int_{- \infty}^{\infty} e^{-t^2}dt$ This is a not mandatory exercise I am struggling with from my Analysis II Class, at the very end of it I am supposed to compute $$\int_{-\infty}^\infty e^{-t^2}dt \tag{*}$$
The most famous way to do that is switch to polar coordinates but I am not supposed to do it that way, one of my tutors even called it a 'dirty trick'.
Instead the exercise gives us the following two Integrals to work with: 

$$F(x)= \int_0^1 \frac{e^{-x^2(1+t^2)}}{1+t^2}dt \text{ and } G(x)= \left( \int_0^x e^{-t^2}dt\right)^2, \ \forall x \in \mathbb{R} $$

I yet fail to see how these two integrals relate to (*) but I can naively follow the instructions given by exercise sheet and hope that the result will follow:

Problem: Show that $F,G$ are of Class $C^1(\mathbb{R})$, compute $F',G'$ for all $x \in \mathbb{R}$ and show that $F+G$ is constant

My approach: There isn't much to do really rather than to quote some Lemmas, clearly the parametric integral $F$ is of class $C^1$ because $f(x,t): \mathbb{R} \times [0,1] \to \mathbb{R}$ given by $f(x,t)=e^{-x^2(1+t^2)}/(1+t^2)$ is of class $C^1$, I can therefore switch the partial derivative with the integral and obtain that  $$F'(x) =-2 x \int_0^1e^{-x^2(1+t^2)}dt$$ I doubt that I can further simplify this integral.
With a little help of the fundamental theorem of Calculus one obtains that $$G'(x)= 2 \left( \int_0^x e^{-t^2}dt\right)e^{-x^2} $$

At this point I hope that the exercise (or my computation) is flawed, at least the limit points of the integrals because I wouldn't know how to show that $F'+G'=0$ for all $x\in \mathbb{R}$ to conclude that $F+G$ is constant. 
The only sense I could make out of $F'$ would be if I integrated it with respect to $x$ rather to $t$. Are there any further simplifications I can do?
 A: $$F(x) = \int_0^1 dt \frac{e^{-x^2 (1+t^2)}}{1+t^2} \implies F'(x) = -2 x e^{-x^2} \int_0^1 dt \, e^{-x^2 t^2} $$
$$G(x) = \left ( \int_0^x dt \, e^{-t^2} \right )^2 = x^2 \left ( \int_0^1 dt \, e^{-x^2 t^2} \right )^2 \\ \implies G'(x) = 2 x \left ( \int_0^1 dt \, e^{-x^2 t^2} \right ) \left [\int_0^1 dt \, e^{-x^2 t^2} - 2 x^2 \int_0^1 dt \,t^2  e^{-x^2 t^2} \right ]$$
Then
$$F'(x) + G'(x) = 2 x \left ( \int_0^1 dt \, e^{-x^2 t^2} \right )\left [-e^{-x^2} + \int_0^1 dt \,(1-2 x^2 t^2)  e^{-x^2 t^2} \right ] $$
Then note that
$$(1-2 x^2 t^2)  e^{-x^2 t^2} = \frac{d}{dt} \left ( t e^{-x^2 t^2} \right )$$
and you'll find that $F'(x) + G'(x) = 0$ identically.  Thus $F(x) = G(x)$ equals a constant, say $F(0)+G(0)$, which is
$$\int_0^1 \frac{dt}{1+t^2} = \frac{\pi}{4} $$
Now take the limit as $x \to \infty$, in which limit $F(x) \to 0$ and thus
$$\lim_{x \to \infty} \left ( \int_0^x dt \, e^{-t^2} \right )^2 = \left ( \int_0^{\infty} dt \, e^{-t^2} \right )^2 = \frac{\pi}{4} $$
A: Hint: In your expression for $F'$, try to write $x^2 (1+t^2)$ as $x^2 + (xt)^2$, use the law $e^{a+b} = e^a e^b$, move the constant term (that does not depend on $t$) outside of the integral, and change variables with $s = xt$.
A: Another "dirty trick" to evaluate this integral, but without the use of polar coordinates, would be the following. Once you accept the use of Fubini's Theorem --- I presume this is actually the dirty trick which your tutor wants to avoid! --- to get to $$(\int_{-\infty}^\infty e^{-x^2}dx)^2=\int_{-\infty}^{\infty}\int_{-\infty}^\infty e^{-( x^2+y^2)}dxdy=:V,$$
one can proceed as follows. The double integral on the right hand side measures the finite volume $V$ of the (infinite) body  in $\mathbb{R}^3$ lying between the $x$-$y$-plane and the graph of the function $(x,y)\mapsto e^{-(x^2+y^2)}$. However the latter surface can be viewed as being obtained by rotation around the $z$-axis of the graph of the function $$[0,\infty)\to(0,1],\qquad x\mapsto z=e^{-x^2}.$$ Viewed as a function of $z$ this is just the graph of the function $$(0,1]\to [0,\infty),\quad z\mapsto x=\sqrt{-\log(z)}, $$ still being rotated around the $z$-axis.
Hence the  volume $V$ can be calculated just as the volume of a body obtained by rotation of a function graph around the $z$-axis is usually calculated:
$$V=\int_0^1x^2\pi dz=-\pi\int_0^1\log( z) dz=-\pi[z\log(z)-z]_0^1=\pi.$$
(I omit the detail that in this integral the lower bound $0$ has to be repalced by $\epsilon > 0$ and then letting $\epsilon$ tend to $0$.)
