# 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.

• Those are the same integral: $d\to d-2$ turns the second into the first. So half your problem is trivial. Feb 28, 2020 at 17:32
• Ha! I never noticed that. Feb 28, 2020 at 17:34
• Have you used Beta function formula? Feb 28, 2020 at 17:39
• See section 8.2.1 here. Feb 28, 2020 at 17:43
• en.wikipedia.org/wiki/Pochhammer_contour Feb 28, 2020 at 20:56

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 $$I = \frac{1}{2} \Gamma\left( \frac{d}{2} \right) \Gamma\left(2 - \frac{d}{2} \right).$$