the integral of $\sin(z^2) \exp\left({-4z^2xy \over y^2-x^2}\right)$ can be written in Fresnel integrals? $$\int_{0}^{(y^2-x^2)/ 4t}s^{-1/2} \sin(s) \exp\left({-4sxy \over y^2-x^2}\right)\mathrm{d}s=2\int_{0}^{\sqrt{(y^2-x^2)/4t}}\sin(z^2) \exp\left({-4z^2xy \over y^2-x^2}\right)\mathrm{d}z$$
I applied substitution of $s=z^2$. My question is how we can write more simple and is it possible to rewrite this integral with respect to Fresnel integrals?
Thanks! 
 A: Yes indeed: 
Rewriting the $\sin(z^2)$ Term by the help of Eulers formula, we get
$$
I(y,x,t)=\frac{1}{2i}\int_0^{\beta(x,y,t)}(e^{i z^2 }-e^{-i z^2 })e^{-\alpha(x,y)z^2}\mathrm{d}z
$$
with $\alpha(x,y)=\frac{4xy}{y^2-x^2}$ and $\beta(x,y,t)=\frac{y^2-x^2}{4t}$.
By a simple rescaling and by using the definition of the Error function $\text{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-x^2}\mathrm{d}x$ we obtain
$$
I(\alpha, \beta)=\frac{\sqrt{\pi}}{4i}\left( \frac{\text{erf}(\beta\sqrt{\alpha-i})}{\sqrt{\alpha-i}}-\frac{\text{erf}(\beta\sqrt{\alpha+i})}{\sqrt{\alpha
+i}}\right)
$$
It is well known and also straightforward (but a little bit tedious) how to express the Error functions in terms of Fresnel integrals:
$$
\text{erf}[x]=(1+i)\left( C\left(\frac{(1-i)x}{\sqrt{\pi}}\right)-iS\left(\frac{(1-i)x}{\sqrt{\pi}}\right)\right)
$$
Where $C(x),S(x)$ denote the two Fresnel integrals. Setting $x=\beta\sqrt{\alpha\pm i}$ we arrive at
$$
I(\alpha,\beta)=\sqrt{\pi}\frac{1+i}{4i}\left[{1 \over \sqrt{\alpha-i}}\left( C\left(\frac{(1-i)\beta\sqrt{\alpha-i}}{\sqrt{\pi}}\right)-iS\left(\frac{(1-i)\beta\sqrt{\alpha-i}}{\sqrt{\pi}}\right)\right)-{1 \over \sqrt{\alpha+i}}\left( C\left(\frac{(1-i)\beta\sqrt{\alpha+i}}{\sqrt{\pi}}\right)-iS\left(\frac{(1-i)\beta\sqrt{\alpha+i}}{\sqrt{\pi}}\right)\right)\right]
$$
which is by the way a much uglier results then just using Error functions...
Furthermore please note that i don't really take care of convergence issues (at $x\rightarrow y$). Please let me know if you want to elaborate more on this point.  
