Calculate $\int_0^1 \ \int_0^1 \ x \sin \lvert x^2-y^2 \lvert \; dx \; dy$ $$\int_0^1 \ \int_0^1  \ x \ \sin \lvert x^2-y^2 \lvert dx \ dy $$
$$\int_0^1 \frac{1}{2} \Big[ \sin \lvert x^2-y^2 \lvert      \Big]_0^1 \ dy= 
\int_0^1 \frac{1}{2} \Big( \sin \lvert 1-y^2 \lvert - \sin \ (y^2) \Big) \ dy  $$
How can I continue the integration?
Thanks!
 A: Ind: you can suppose that:  $**x=cosh(u)**$ and $**y=sinh(v)**$  
A: First of all, $y^2\leq 1$, so you actually have $\sin(1-y^2)$. Then you have to use $$\int\sin(y^2)dy=\sqrt{\frac{\pi}{2}}{\rm Si}\left(\sqrt{\frac{2}{\pi}}y\right)$$ twice and $$\int\cos(y^2)dy=\sqrt{\frac{\pi}{2}}{\rm Ci}\left(\sqrt{\frac{2}{\pi}}y\right),$$ in which ${\rm Si}$ and ${\rm Ci}$ are the integral sine and cosine, respectively.
A: The integral is equal to
$$\begin{align}I &=\int_0^1 dx \, x \, \int_0^x dy \, \sin{(x^2-y^2)} - \int_0^1 dx \, x \, \int_x^1 dy \, \sin{(x^2-y^2)}\\ &= \int_0^1 dx \, x \, \int_0^x dy \, \sin{(x^2-y^2)} - \int_0^1 dy \, \int_0^y dx \, x \, \sin{(x^2-y^2)} \\ &= \int_0^1 dx \, x \, \int_0^x dy \, \sin{(x^2-y^2)} + \int_0^1 dx \, \int_0^x dy \, y \, \sin{(x^2-y^2)} \\ &= \int_0^1 dx \, \int_0^x dy \, (x+y) \sin{(x^2-y^2)} \\ &= \frac12 \int_0^1 dv \, \int_0^{2-v} du \, u \sin{(u v)}\\ &= \frac12 \int_0^1 dv \left [\frac{\sin{[v (2-v)]}}{v^2} - \frac{\cos{[v (2-v)]}}{v} \right ] \end{align}$$
which finally, is expressible in terms of the Fresnel integrals
$$C(z) + i S(z) = \int_0^z dt \, e^{i \pi t^2/2}$$
as

$$I = 2 - \sin{1} - \sqrt{2 \pi} \left [C \left (\sqrt{\frac{2}{\pi}} \right ) \cos{1}+ S \left (\sqrt{\frac{2}{\pi}} \right ) \sin{1} \right ] $$

Here is a summary the steps I took:
1) Split up integration region according to whether $x^2-y^2$ is pos or neg
2) Reverse order of integration in second double integral
3) Switch $y$ and $x$
4) Combine the double integrals into a single double integral
5) Change coordinates to $u=x+y$, $v=x-y$.  factor of $1/2$ in front is the Jacobian of the coordinate transformation.
6) Integrate wrt $u$
