Differentiation under the integral sign — where is my mistake? I'm trying to find 
$$\int_0^\infty \sin \left( x^2 \right)\,dx$$ 
by the method of differentiation under the integral sign. The idea is to use differentiation with respect to $t$ on $A(t)$ — defined below — and then let $t$ approach infinity and take the square root to find the Fresnel integral. 
Let 
$$A(t) = \left( \int_0^t \sin(x^2)\,dx \right)^2$$
$$A'(t) = 2\sin(t^2) \int_0^t \sin(x^2)\,dx$$
Let $x=yt$
$A'(t) = 2\sin(t^2) \int_0^1 \sin(t^2y^2)t\,dy $
$A'(t) = \int_0^1 2t\sin(t^2)\sin(t^2y^2)\,dy $
$A'(t) = \int_0^1 t(\cos(t^2-t^2y^2)-\cos(t^2+t^2y^2))\,dy $
$A'(t) = \int_0^1 t(\cos(t^2(1-y^2))-\cos(t^2(1+y^2)))\,dy $
$A'(t) = \frac{1}{2}\int_0^1 \frac{\partial}{\partial t} (\frac{\sin(t^2(1-y^2))}{1-y^2}-\frac{\sin(t^2(1+y^2))}{y^2+1})\,dy $
$A'(t) = \frac{1}{2}\frac{\text{d}}{\text{d}t}\int_0^1 \frac{\sin(t^2(1-y^2))}{1-y^2}-\frac{\sin(t^2(1+y^2))}{y^2+1} \, dy $
By the Fundamental Theorem of Calculus:
$\int A'(t)\,dt + C = A(t)$
If we take the limit as $\lim_{t\to 0}$:
EDIT: $\lim_{t\to 0} \int{A'(t)dt} + C = \lim_{t\to 0}A(t)$
$\lim_{t\to 0} \int{A'(t)dt} = 0$ and $\lim_{t\to 0}A(t) = 0$
So $0 + C = 0$ and $C=0$
Thus, $\int A'(t)\,dt = A(t)$,
But, if we take the limit as $\lim_{t\to \infty}$:
EDIT: $\lim_{t\to \infty} \int{A'(t)dt} = \lim_{t\to \infty}A(t)$
I haven't been able to confirm it, but I am pretty sure from numerical calculations that $\lim_{t\to \infty} \int{A'(t)dt} = 0$. 
But we know that $\lim_{t\to \infty}A(t)$ should be $\pi/8$.
And $0\neq \pi/8$.
I may have made a simple algebra or calculus mistake, but I haven't caught it. 
You help is very much appreciated. 
 A: I stopped reading when I got to the first mistake. When you set $x=ty$, you forgot to put $dx=tdy$ in your integral.
A: You cannot use $\lim\limits_{t\to\infty}A'(t)=0$ to conclude $\lim\limits_{t\to\infty}A(t)=0$. In fact, something of the opposite is true: for any limit $\lim\limits_{t\to\infty} A(t)=L$ of a nice function $A$, its existence implies $\lim_{t\to\infty}A'(t)=0$, no matter what the original limit $L=\lim\limits_{t\to\infty}A(t)$ is. Indeed, adding a constant to the function $A(t)$ will alter the value of the limit $L=\lim\limits_{t\to\infty}A(t)$ but it will not change the fact that $\lim\limits_{t\to\infty}A'(t)=0$.
It is also not true that analyzing the behavior of a function $A(t)$ and its derivative $A'(t)$ near the value $t=0$ will tell us what happens for $A(t)$ as we take $t\to\infty$. For instance, $A(t)=A'(t)=0$ is completely possible even if $A(t)$ is increasing on the interval $(0,\infty)$. This is a local-global distinction: what happens locally does not generally what happens globally. There are certain exceptions, for instance with an analytic function, but even with analytic functions we would have to specify all of the values $A(0)$, $A'(0)$, $A''(0)$, $A'''(0)$, $\cdots$ in order to force any determination of what happens to $A(t)$ as  we take $t\to\infty$.
In particular, suppose we accept the logic that $A(0)=A'(0)=0$ implies $A(t)\to0$ as $t\to\infty$, which we know is not true by considering for instance $A(t)$ increasing on $(0,\infty)$. If we define the function $A(t)=\int_0^t f(\tau)d\tau$ for any nice function $f$ for which $\int_0^\infty f(\tau)d\tau$ converges and $f(0)=0$, this logic would imply that any and all integrals $\int_0^\infty f(\tau)d\tau$ (with $f(0)=0$) are zero!
If you want an explicit example, consider $f(\tau)=\tau^2e^{-\tau}$. Then $f(0)=0$, and $A(0)=0$ but $A(t)=\int_0^t \tau^2e^{-\tau}d\tau\to \Gamma(3)=2$ as $t\to\infty$.
A: I don't think your approach will work. But if you know that 
$\int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt \pi}{2}$, then you can do a contour integral 
$\int _{c}e^{-z^{2}}dz$, where $c$ is the line $0$ to $r$, then counterclockwise from $r$ to $re^{i\pi /4}$, and finally back along the line from $re^{i\pi /4}$ to $0$. This is a closed contour and the integrand is analytic inside $c$, so the integral must be $0$. Calculating the integrals along the pieces of the contour, letting $r\rightarrow \infty$ will give you the result. 
