Integral $f(t)=\int_0^{\pi/2}\sin(t\tan x)dx$ 
Evaluate the integral
  $$f(t)=\int_0^{\pi/2}\sin(t\tan x)dx.$$

I came across this integral after having shown that its counterpart can be evaluated exactly:
$$\int_0^{\pi/2}\cos(t\tan x)dt=\frac{\pi}{2}e^{-|t|}.$$
I figured that $f$ would be as easy but it is actually presenting me with some difficulties. I start off by using the substitution $\tan x\mapsto x$:
$$f(t)=\int_0^\infty \frac{\sin tx}{1+x^2}dx,$$
then, taking the Laplace transform:
$$\begin{align}
\mathcal{L}\{f\}(s)&=\int_0^\infty \frac1{1+x^2}\int_0^\infty e^{-st}\sin(xt)dtdx\\
&=\int_0^\infty \frac1{1+x^2}\text{Im}\left[\int_0^\infty e^{-(s-ix)t}dt\right]dx\\
&=\int_0^\infty \frac1{1+x^2}\frac{x}{s^2+x^2}dx\\
&=\frac12\int_0^\infty \frac{dx}{(1+x)(s^2+x)}\\
&=\frac1{2s^2-2}\int_0^\infty \left[\frac{1}{1+x}-\frac{1}{s^2+x}\right]dx\\
&=\frac{\ln s}{s^2-1}.
\end{align}$$
Now, I see that $f(t)=\mathcal{L}^{-1}\left\{\frac{\ln s}{s^2-1}\right\}(t),$ but I do not know how to compute that inverse Laplace transform. Could I have some help?
 A: Mathematica gives:
$$\frac{1}{4} \sqrt{\pi } t G_{1,3}^{2,1}\left(\frac{t^2}{4}|
\begin{array}{c}
 0 \\
 0,0,-\frac{1}{2} \\
\end{array}
\right)$$
involving the Meier G function.
A: After Kemono Chen's comment.
Starting with 
$$f(t)=\int_0^\infty \frac{\sin (tx)}{1+x^2}\,dx=\int_0^\infty \frac{\sin (tx)}{(x+i)(x-i)}\,dx$$
$$f(t)=\frac i 2 \left(\int_0^\infty \frac{\sin (tx)}{x+i}\,dx -\int_0^\infty \frac{\sin (tx)}{x-i}\,dx\right)$$
Consider now
$$I(a)=\int  \frac{\sin (tx)}{x+a}\,dx=\int\frac{\sin (t (y-a))}{y}\,dy$$
$$I(a)=\cos(at) \int \frac{\sin (ty)}{y}\,dy-\sin(at)\int \frac{\cos (ty)}{y}\,dy$$ Now, make $z=ty$
$$I(a)=\cos(at) \int \frac{\sin (z)}{z}\,dz-\sin(at)\int \frac{\cos (z)}{z}\,dz=\cos (a t)\,\text{Si}(z) -\sin (a t)\,\text{Ci}(z) $$
Back to $x$
$$I(a)=\cos (a t)\, \text{Si}(t(x+a))-\sin (a t)\, \text{Ci}(t(x+a))$$ Assuming $t >0$
$$J(a)=\int_0^\infty  \frac{\sin (tx)}{x+a}\,dx=\text{Ci}(a t) \sin (a t)+\frac{1}{2} (\pi -2 \text{Si}(a t)) \cos (a t)$$
$$J(a)-J(-a)=(\text{Ci}(-a t)+\text{Ci}(a t)) \sin (a t)-2 \text{Si}(a t) \cos (a t)$$
$$f(t)=\frac i 2 \left(J(i)-J(-i)\right)=\text{Shi}(t) \cosh (t)-\text{Chi}(t) \sinh (t)$$ which is Kemono Chen's result.
