Dimensional regularization integral I'm new here.  Are there places to put specific problems or do they just go to a general list?
There are some similar problems posted, but they are all a bit different and I can't see how to use the advice.
I have two integrals I am having troubles with:
$$\int_0^{ \infty } dk ~ k^{d - 1} \dfrac{1}{(k^2 + v^2)^2} = \dfrac{1}{2} (v^2)^{d/2 -2} \Gamma \left ( \dfrac{d}{2} \right ) \Gamma \left ( 2 - \dfrac{d}{2} \right )$$
and
$$\int_0^{ \infty } dk ~ k^{d - 1} \dfrac{k^2}{(k^2 + v^2)^2} = \dfrac{1}{2} (v^2)^{d/2 - 1} \Gamma \left ( 1 + \dfrac{d}{2} \right ) \Gamma \left ( 1 - \dfrac{d}{2} \right ).$$
Perhaps I'm not being clever enough. I've tried substitution, series methods, taking the derivative of the integral wrt v, and contour integration.  All I can seem to get is that, for most integer values of d, that the integrals do not converge.  I'd appreciate a guide to the solution, but I'll settle for someone telling me what the method is.
 A: As pointed out in the comments, the two integrals are really the same. Also, just by redefining $k \rightarrow v k$ in the integral, we can easily scale out the $v$-dependence:
$$
\int_0^{ \infty } dk ~ k^{d - 1} \dfrac{1}{(k^2 + v^2)^2} = (v^2)^{d/2 -2} \int_0^{ \infty } dk ~ k^{d - 1} \dfrac{1}{(k^2 + 1)^2},
$$
so it suffices to solve the $v=1$ case.
All we will need is the definition of the gamma function:
$$
\Gamma(z) = \int_0^{\infty} d \lambda \, \lambda^{z - 1} e^{-\lambda}.
$$
This formula immediately implies the so-called Schwinger parametrization of your integrand, which is often useful for loop diagrams:
$$
\frac{1}{A^{z}} = \frac{1}{\Gamma(z)}\int_0^{\infty} d \lambda \, \lambda^{z - 1} \, e^{- \lambda A},
$$
which follows from simply scaling $\lambda \rightarrow A \lambda$ in the definition of the gamma function. Using the $z = 2$ case of this formula, your integral becomes
$$
I = \int_0^{ \infty } dk ~ k^{d - 1} \dfrac{1}{(k^2 + 1)^2} = \int_0^{ \infty } dk \int_0^{ \infty } d\lambda ~ \lambda \, k^{d - 1} e^{-\lambda(k^2 + 1)}.
$$
Now we substitute $\alpha = \lambda k^2$, and after some algebra, we get
$$
I = \frac{1}{2} \int_0^{ \infty } d\alpha \int_0^{ \infty } d\lambda ~ \lambda^{1 - d/2} \, \alpha^{d/2 - 1} e^{-\alpha} e^{-\lambda}.
$$
Now both integrals are precisely in the form of the gamma function, so we have
$$
I = \frac{1}{2} \Gamma\left( \frac{d}{2} \right) \Gamma\left(2 - \frac{d}{2} \right).
$$
