Evaluating $\int_0^\infty \frac{\sin{(tx)}}{x(x^2+1)}dx$

Question

I want to show that $\displaystyle\int_0^\infty\frac{\sin{(tx)}}{x(x^2+1)}dx=\frac{\pi}{2}(1-e^{-t})\tag{1}$ for $t>0 $ by justifying differentiation under the integral sign and using $\displaystyle\int_R \frac{\cos{(tx)}}{x^2+1}=\pi e^{-|t|}$.

My attempt:- Let $F(t)=\displaystyle\int_0^\infty \frac{\sin{(tx)}}{x(x^2+1)}dx$. If we try to compute $F'(t)$ for $t>0$ using differentiation under the integral sign, then we get

$$F'(t)=\displaystyle\int_0^\infty \frac{\partial}{\partial t}\left(\frac{\sin{(tx)}}{x(x^2+1)}\right)\,dx=\displaystyle\int_0^\infty \frac{\cos{(tx)}}{x^2+1}\,dx=\frac{\pi}{2}e^{-t}.\tag{2}$$

Now how to answer this question?

My hp 50g calculator unable to give answer using 'RISCH' and 'INTVX' commands If any member knows the answer may reply with correct answer.

Answer 1

For $t>0$, 2) gives $F(t)=\frac {\pi} 2 (1-e^{-t})$ by integration and the condition $F(0)=0$. (For $t <0$ you can use the fact that LHS of 1) is an odd function).