Let $f(x)=\int_{0}^{x}\frac{\sin(t)}{t^{3/2}}dt$ find $\lim_{k \to \infty} \int_{0}^{\infty}k^{3/2}f(x)e^{-kx} dx$ Let $f(x)=\int_{0}^{x}\frac{\sin(t)}{t^{3/2}}dt$ find $\lim_{k \to \infty} \int_{0}^{\infty}k^{3/2}f(x)e^{-kx} dx$
I am not sure how to approach the problem. I tried integrating by parts and using Fubini (2 different tries), but both came down to some integrals that I could not evaluate.
Notes:
$f$ is absolutely continuous. and pointwise the sequence inside converges to $0$. I tried to find a dominating function (like $f(x)e^{-x}$)
 A: In other terms, you are trying to understand the behaviour of the Laplace transform  of
$$ f(x)=\int_{0}^{x}\frac{\sin(t)}{t\sqrt{t}}\,dt.$$
The Laplace transform of $\frac{\sin(t)}{t\sqrt{t}}$ is $\frac{\sqrt{2\pi}}{\sqrt{s+\sqrt{s^2+1}}}$ (proof below), so the Laplace transform of $f$ is 
$$ \frac{\sqrt{2\pi}}{s\sqrt{s+\sqrt{s^2+1}}} $$
and the wanted limit equals $\color{red}{\sqrt{\pi}}$.

By the self-adjointness of the Laplace transform, under weak regularity assumptions we have $\int_{0}^{+\infty}f(x)g(x)\,dx = \int_{0}^{+\infty}(\mathcal{L}f)(x)\cdot(\mathcal{L}^{-1}g)(x)\,dx$, hence
$$ \int_{0}^{+\infty}\sin(t)e^{-st}\frac{dt}{t\sqrt{t}} = \int_{0}^{+\infty}\frac{1}{1+(s+u)^2}\cdot\frac{2\sqrt{u}}{\sqrt{\pi}}\,du $$
equals, via $u\mapsto v^2$,
$$ \frac{4}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{v^2}{1+(s+v^2)^2}\,dv 
\stackrel{v\mapsto z\sqrt{s}}{=} \frac{4\sqrt{s}}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{sz^2}{1+s^2(1+z^2)^2}\,dz $$
which can be computed through many standard techniques. Even without explicit computations, we may see that the wanted limit equals
$$\frac{4}{\sqrt{\pi}}\lim_{s\to +\infty}\int_{0}^{+\infty}\frac{z^2}{\frac{1}{s^2}+(1+z^2)^2}\,dz \stackrel{\text{DCT}}{=} \frac{4}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{z^2}{(1+z^2)^2}\,dz=\color{red}{\sqrt{\pi}}.$$
