How can I attack $\int_{-\infty}^\infty \frac{x}{[(x-vt)^2+a^2]^2} dx$? I want to evaluate
$$\int_{-\infty}^\infty \frac{x}{[(x-vt)^2+a^2]^2} dx$$
I know that 
$$\int \frac{1}{(\xi^2+a^2)^2} = \frac{\xi}{2a^2(\xi^2+a^2)}+\frac{1}{2a^3}\arctan\frac{\xi}{a}$$
Using integration by parts and choosing $a:= a$, $\xi := x-vt$ and thus
$$u = x \Rightarrow u' = 1  $$
$$g' = \frac{1}{[(x-vt)^2+a^2]^2} \Rightarrow g = \frac{x-vt}{2a^2((x-vt)^2+a^2)}+\frac{1}{2a^3}\arctan\frac{x-vt}{a}$$
Therefore I get 
$$\int_{-\infty}^\infty \frac{x}{[(x-vt)^2+a^2]^2} dx = \left. x\cdot \frac{x-vt}{2a^2((x-vt)^2+a^2)}+\frac{1}{2a^3}\arctan\frac{x-vt}{a}\right\vert_{-\infty}^\infty - \int_{-\infty}^\infty \frac{x-vt}{2a^2((x-vt)^2+a^2)}+\frac{1}{2a^3}\arctan\frac{x-vt}{a} dx$$
Which isn't really helpful especially because 
$$\left.x\cdot \frac{x-vt}{2a^2((x-vt)^2+a^2)}+\frac{1}{2a^3}\arctan\frac{x-vt}{a}\right\vert_{-\infty}^\infty$$
Makes me divide $\frac{\infty}{\infty}$ in the frist term ...
So how can I do this better?
Thank you very much for your help.
FunkyPeanut
 A: Start by substituting $\xi=x-vt$ to get
$$\eqalign{\int_{-\infty}^\infty \frac{x}{((x-vt)^2+a^2)^2} dx
  &=\int_{-\infty}^\infty \frac{\xi+vt}{(\xi^2+a^2)^2} dx\cr
  &=\int_{-\infty}^\infty \frac{\xi}{(\xi^2+a^2)^2} dx
    +vt\int_{-\infty}^\infty \frac{1}{(\xi^2+a^2)^2} dx\ .\cr}$$
Now the first integral is zero because the integrand is odd (but see below), and you already know how to do the second, giving
$$vt\biggl[\frac{\xi}{2a^2(\xi^2+a^2)}+\frac{1}{2a^3}\arctan\frac{\xi}{a}\biggr]_{-\infty}^\infty\ .$$
Now if $\xi\to\pm\infty$, the first term tends to $0$, because it has $\xi$ in the numerator but $\xi^2$ in the denominator.  Also, the $\arctan$ term tends to $\pm\pi/2$, provided that $a$ is positive.  So it all comes out to
$$\int_{-\infty}^\infty \frac{x}{((x-vt)^2+a^2)^2} dx
  =\frac{vt\pi}{2a^3}\ .$$
Note that if $f$ is an odd function,
$$\int_{-\infty}^\infty f(\xi)\,dx=0\ ,$$
provided that $\int_0^\infty f(\xi)\,dx$ converges.  This is true in your problem because the integrand is effectively $1/\xi^3$ for large $\xi$.
