$\int_{-\infty}^{\infty} e^{tx} \frac{1}{\pi(1+x^2)} \mathrm dx$ which ways to compute this integral:
$$
\int_{-\infty}^{\infty} e^{tx} \frac{1}{\pi(1+x^2)} \mathrm dx
$$
for different cases of t. t can be any value in $\mathbb{R}$.  I was stuck at how to get the antiderivative of the integrand. Thank you!
 A: When $t \neq 0$, the first integral diverges and the second one converges only if $t < 0$. In the latter case, if you call your integral $f(t)$, then differentiating under the integral sign gives that 
$$f''(t) + f(t) = \int_0^{\infty} {e^{tx} + x^2e^{tx} \over \pi(1 + x^2)}$$
$$ = {1\over \pi} \int_0^{\infty}e^{tx}$$
The latter integral integrates to ${\displaystyle-{1 \over t}}$, so that you have ${\displaystyle f''(t) + f(t) = -{1 \over \pi t}}$. This can be solved via variation of parameters, and (according to wolframalpha at least) it does not have an elementary expression. 
On the other hand the nonelementary functions in the solution just involve indefinite integrals of ${\displaystyle {\sin(t) \over t}}$ and ${\displaystyle{\cos(t) \over t}}$, so if you want to do something this advanced you might be able to get something satisfactory using initial conditions for this differential equation. 
A: First, check whether they converge.  You will find that the first one doesn't unless $t=0$, and the second one doesn't unless $t\leq0$.  For the first integral, when $t=0$, you can use the arctangent function.  For the second, when $t\lt0$, you can try to use complex contour integration methods as seen in many texts on complex analysis.
Edit: I was hasty in posting this, and hadn't actually thought through how the contour integration would go.  Based on the other answers, it looks like it wouldn't be so straightforward.
A: Correspondingly to Zaricuse's second paragraph, according to this (see the formula before the last one), for any $s > 0$,
$$
\int_0^\infty  {e^{ - sx} \frac{1}{{\pi (1 + x^2 )}}\,{\rm d}x}  = \frac{1}{\pi } \Big \lbrace \cos (s)\Big[\frac{\pi }{2} - {\rm Si}(s)\Big] - \sin (s){\rm Ci}(s)\Big \rbrace,
$$
where ${\rm Si}(s) = \int_0^s {\frac{{\sin u}}{u}} \,{\rm d}u$ is the Sine Integral, and ${\rm Ci}(s) = \int_s^\infty  {\frac{{\cos u}}{u}\,{\rm d}u}$ the Cosine Integral.
EDIT: Probabilistic interpretation of the integrals.
The function $f(x) = \frac{1}{{\pi (1 + x^2 )}}$, $x \in \mathbb{R}$, is the density function of a standard Cauchy random variable $X$. The divergent integral
$$
{\rm E}[e^{tX}] = \int_{ - \infty }^\infty  {e^{tx} f(x)\,{\rm d}x}  = \infty, \;\; t \neq 0, 
$$
corresponds exactly to the (elementary) fact that the moment-generating function of the Cauchy distribution does not exist.
On the other hand, the characteristic function is well known to be
$$
{\rm E}[e^{{\rm i}{\rm t}X}] = \int_{ - \infty }^\infty  {e^{{\rm i}tx} f(x)\,{\rm d}x}  = e^{ - |t|}, \;\; t \in \mathbb{R}.
$$
Finally, the function $\tilde f(x) = \frac{2}{{\pi (1 + x^2 )}}$, $x > 0$, is the density function of $|X|$. Then, the Laplace transform of $|X|$ is 
$$
{\rm E}[e^{-t|X|}] = \int_0^\infty  {e^{ - tx} \tilde f(x)\,{\rm d}x},\;\; t > 0,
$$
and, according to the first paragraph, can be expressed in terms of the Sine Integral and Cosine Integral functions.
