Clever substitution on evaluating $\int_{0}^{\infty} \frac{\sin(2x+\frac{x^2}{t^2})}{2x+\frac{x^2}{t^2}} \, dx$ I've been working on evaluating some integrals that come up when studying Fresnel integrals, and I've come across this one: $$\int_{0}^{t^2}  \frac{\sin(2x+\frac{x^2}{t^2})}{2x+\frac{x^2}{t^2}} \, dx$$ and immediately it reminded me of the famous integral:
$$\int_{0}^{\infty} \frac{\sin{x}}{x} \, dx = \frac{\pi}{2}$$ with which I dared to conjecture that:
$$\lim_{t\rightarrow \infty}{\int_{0}^{t^2}  \frac{\sin(2x+\frac{x^2}{t^2})}{2x+\frac{x^2}{t^2}} \, dx}=\frac{\pi}{4}$$
I have tried to do substitutions without success, I also tried to use something like the dominated convergence theorem to swap the integral with the limit and I think it may work, but, will there be some substitution that allows me to evaluate the integral without using something like the dominated convergence theorem ?
Thank you all

Since $t$ is large, use the expansion
$$ \frac{\sin(2x+\frac{x^2}{t^2})}{2x+\frac{x^2}{t^2}}\sim \frac{\sin (2 x)}{2 x}$$
$$\int_{0}^{a} \frac{\sin(2x)}{2x} \, dx=\frac{\text{Si}(2 a)}{2}\quad \to \quad \frac \pi 4$$  as you conjectured.
Edit
If you make
$2x+\frac{x^2}{t^2}=y$ the integral becomes
$$\int_0^{2t^2+1}\frac{t \sin (y)}{2 y \sqrt{t^2+y}}\,dy$$ Expanded as a series for large values of $t$, the integrand is
$$\frac{\sin (y)}{2 y}-\frac{\sin (y)}{4 t^2}+\frac{3 y \sin (y)}{16
   t^4}+O\left(\frac{1}{t^6}\right)$$ Integrating termwise
$$I=\frac{1}{2} \text{Si}\left(2 t^2+1\right)-\frac{\sin
   ^2\left(t^2+\frac{1}{2}\right)}{2 t^2}+\frac{3 \left(\sin \left(2
   t^2+1\right)-\left(2 t^2+1\right) \cos \left(2 t^2+1\right)\right)}{16
   t^4}+O\left(\frac{1}{t^6}\right)$$ and then the result.
