Regarding this post
I would like to ask the community for hints regarding this problem. Since the steps done on that original post, did not yield the correct result, I have done a different approach.
\begin{equation} \int_0^\infty \frac{\sin t}{\sqrt{t}}dt=\frac{1}{2i}\int_0^\infty \frac{e^{it}-e^{-it}}{\sqrt{t}}dt \end{equation}
for simplicity we split in two integrals, and solve the first here, which gives the solution to the second too
\begin{equation} \frac{1}{2i}\int_0^\infty \frac{e^{it}}{\sqrt{t}}dt=\frac{1}{2i}\int_0^\infty e^{\frac{1}{2}\ln t}e^{it}dt=\frac{1}{2i}\int_0^\infty e^{{\frac{1}{2}\ln t}+it}dt \end{equation}
So we want to solve this now,
\begin{equation} \frac{1}{2i}\int_0^\infty e^{{\frac{1}{2}\ln t}+it}dt \end{equation}
and it is tempting to set $u=\frac{1}{2}\ln t+it$, but then $du=\frac{1}{2t}+i$, so we can't get rid of that $t$ in the substitution.
Alternatively, we can write $e^{it}=z$, and form the complex integral
\begin{equation} \frac{1}{2i}\int_0^\infty \sqrt{t}z dz \end{equation}
But here we need to change the t-variable into a complex variable. So I suggest doing a Möbius transformation on of t, but since it is real, it doesn't apply.
Is there any chance here to solve this by some other way, or use this with some modifications?
The first thing I should do is to simplify the problem using $t=x^2$ $$I=\frac{1}{2i}\int \frac{e^{it}-e^{-it}}{\sqrt{t}}\,dt=-\frac i 2\int \left(e^{i x^2}-e^{-i x^2}\right)\,dx$$ which are quite standard since $$\int e^{a x^2}\,dx=\frac{\sqrt{\pi } }{2 \sqrt{a}}\text{erfi}\left(x\sqrt{a} \right)$$ where appears the imaginary error function (which is not elementary).
So $$I=-i\frac { \sqrt \pi} 4\Bigg[\frac{\text{erfi}\left(x\sqrt{i} \right)}{\sqrt{i}} -\frac{\text{erfi}\left(x\sqrt{-i} \right)}{\sqrt{-i}} \Bigg]$$