integral involving hyperbolic tangent and cosine I am looking for the solution to the following integral
$$\int_{\mathbb{R}} \frac{\tanh(x)\cos(xt)}{x(x^2+1/4+\alpha)}dx \quad\text{with}\quad \alpha, t\in \mathbb{R}$$. 
Any hints at approaching the solution or the solution itself would be much thanked for. 
 A: The first step I would take (if you're doing it by hand) is converting the trig functions to something easier to work with (their exponential definitions):
$\tanh(x) = {e^x - e^{-x}\over e^{x} + e^{-x}}$
$\cos(xt) = {e^{-ixt} - e^{ixt}\over 2}$
Next I would try integration by parts (It's not the $u,v$ form you're used to):
$\int[f(x)g(x)dx] = f(x)\int[g(x) dx] - \int[{d \over dx}[f(x)]\int[g(x) dx] dx]$
However, integrating the trig part alone produces a rather messy result...
https://www.wolframalpha.com/input/?i=integral(tanh(x)*cos(tx))
But, combined with the polynomial in the denominator, and the fact you're integrating over all reals, might have a simpler result. (Wolfram Alpha isn't able to compute your actual integral)
Edit:
Actually, there might be a better method. Still using integration by parts, let:
$f(x) = \tanh(x)$
$g(x) = {\cos(xt) \over x(x^2 + 1/4  + a)}$
Integrating $g(x)$, as seen here:
https://www.wolframalpha.com/input/?i=integral(cos(tx)%2F(x(x%5E2+%2B+(1%2F4)+%2B+a)))
Is a little more workable, because the $Si(x)$ and $Ci(x)$ have nice identities for the limits $-\infty$ and $\infty$:
$Si(-\infty) = -\pi / 2$
$Si(\infty) = \pi / 2$
$Ci(-\infty) = \pi i$
$Ci(\infty) = 0$
Edit:
Okay now I'm invested. Letting:
$A = \sqrt{a + 1/4}$
Then making the $g(x)$ integral neater (don't be scared):
$\int[g(x)dx] = -{1 \over 2A^2}(\cosh(At)(Ci(t(x-Ai)) + Ci(t(x+Ai))) + i\sinh(At)(Si(t(-x+Ai)) + Si(t(x+Ai)))-2Ci(tx))$
However this is indefinite. The definite integral (using our identities) is magically simple:
$\int_{-\infty}^{\infty}[g(x)dx] = {\pi i \over A^2}(1 + \cosh(At))$
This is a constant (with respect to $x$). Now this is where I made a mistake saying you can pull out the $g(x)$ integral. The integral inside the right integral is actually indefinite:
$\int_a^b[f(x)g(x)dx] = f(x)|_a^b\int_a^b[g(x) dx] - \int_a^b[{d \over dx}[f(x)]\int[g(x) dx] dx]$
Where ${d \over dx}[\tanh(x)] = sech^2(x)$
Meaning it's still complicated. Sorry about that.
