Find exact formula of integral function Let $$f(t)=\int_{0}^{\infty}\frac{\sin(tx)dx}{x^2+1}$$

Find exact formula for $f$

Very similar integral $$g(t)=\int_{0}^{\infty}\frac{\cos(tx)dx}{x^2+1}$$, has an exact form. It is $$\frac{\pi e^{-t}}{2}$$
I have tried almost everything, including Laplace transform and Feynman integration.
But what if the answer hides in complex analysis? Will it be simpler to calculate some integral of $\frac{e^{tx}}{x^2+1}$?
EDIT:
The thing i really looking for is to find an integral
$$I=\int_{0}^{\infty}\frac{e^{itx}}{Ax^2+Bx+C}dx$$
Where $A,B,C$ are complex numbers
I was only managed to change it a little
$$I=\int_{0}^{\infty}\frac{e^{itx}}{(x+p)^2+q}dx$$, where $p=\frac{B}{2A}$ and $q = \frac{C}{A}- p^2$
And this is similar to orginal $f$ and $g$
Thank you for any help
Regards
 A: You can do that with the Residue theorem, see the first example here:
https://en.wikipedia.org/wiki/Residue_theorem
EDIT: sorry, I didn't notice that it is actually more difficult than that: for the cosine case you can evaluate the integral as one half of the integral over the real line, but not for the sine case. This makes the choice of the integration path in the residue theorem problematic. The actual answer involves the "exponential integral function" $Ei$, which is an integral function that can be evaluated analytically only for special values. 
See: https://en.wikipedia.org/wiki/Exponential_integral
A: Notice that:
$$f(t)=\int_0^\infty\frac{\sin(tx)}{x^2+1}dx$$
$$f'(t)=\int_0^\infty\frac{x\cos(tx)}{x^2+1}dx$$
$$f''(t)=-\int_0^\infty\left(\frac{1}{x^2+1}-1\right)\sin(tx)dx$$
$$f''(t)=f(t)-\int_0^\infty\sin(tx)dx$$
Now if we use Laplace transform we obtain:
$$(s^2-1)F(s)-(s+1)f(0)=-\int_0^\infty\int_0^\infty\sin(tx)e^{-st}dxdt=-\int_0^\infty\int_0^\infty\sin(tx)e^{-st}dtdx$$
Now first solve the inner integral:
$$I_1=\int_0^\infty\sin(tx)e^{-st}dt=\Im\int_0^\infty e^{-t(s-ix)}dt=\Im\left(\frac{1}{s-ix}\right)=\frac{x}{x^2+s^2}$$
Now the outer integral:
$$I_2=-\int_0^\infty\frac{x}{x^2+s^2}dx=-\frac{1}{2}\lim_{u\to\infty}\ln\left|1+\frac us\right|=-\frac 12\ln(2)$$
Now we have:
$$(s^2-1)F(s)=-\frac 12\ln(2)$$
$$F(s)=-\frac 12\ln(2)\frac{1}{s^2-1}$$
$$f(t)=\frac{\ln(2)}{4}\left(e^{-t}-e^t\right)$$
However I feel I have made an error in evaluating the limit, I presumed the value of $s$ can be manipulated so that the limit converges
