I am doing loop integral in quantum field theory, and an issue in shifting integration variable is giving me a problem. Let me illustrate with an example.

I have an integral that looks approximately as

$$I = \int^1_0 ~\mathrm dx \int_{-\infty}^{\infty} \frac{\mathrm d^dk}{(2\pi)^d}\frac{(k+p)\cdot\gamma}{((k+px)^2+m^2x^2)^2}$$

where $d = 4-2\epsilon$, which is used often in "dimensional regularization" in physics and $\gamma$ is the Dirac gamma matrices also used in physics.

I approach this integral in two different methods:

1) First, I shift $k = k'-px$ and assume $\mathrm d^4k = \mathrm d^4k'$ since integration is from $-\infty $ to $\infty$, I get, $$I = \int^1_0 ~\mathrm dx \int_{-\infty}^{\infty} \frac{\mathrm d^dk}{(2\pi)^d}\frac{(k+p(1-x))\cdot\gamma}{(k^2+m^2x^2)^2}.$$ (By the way, $p^2 = (p\cdot\gamma) (p\cdot\gamma)$).

Now say I integrate to get $I = \displaystyle \int^1_0~\mathrm dx f(x,p\cdot\gamma)$, then take derivative with respect to $p\cdot\gamma$:

$$\frac{\mathrm d}{\mathrm d p\cdot\gamma} I = \int_0^1 ~\mathrm dx\frac{\partial}{\partial p\cdot\gamma}~(f(x,p\cdot\gamma))\, .$$

2) This time, I take the derivative w.r.t., $p\cdot\gamma$ first to get:

\begin{align}\frac{\mathrm d}{\mathrm d p\cdot\gamma} I &=\int^1_0 ~\mathrm dx \int_{-\infty}^{\infty} \frac{\mathrm d^dk}{(2\pi)^d}\frac{\partial}{\partial p\cdot\gamma}\frac{(k+p)\cdot\gamma}{((k+px)^2+m^2x^2)^2}\\ &=\int^1_0 ~\mathrm dx \int_{-\infty}^{\infty} \frac{\mathrm d^dk}{(2\pi)^d}\left(\frac{1}{((k+px)^2+m^2x^2)^2}+\frac{((k+p)\cdot\gamma)(2x(k+px)\cdot\gamma)}{((k+px)^2+m^2x^2)^3}\right)\end{align}

Now, I shift $k=k'-px$ again, and I get a different answer.

Why are they different from each other, and if I want to get $\frac{\mathrm d}{\mathrm d p\cdot\gamma} I$, which one should I use? I would assume that second method is correct, if there is difference in answer, but all the textbooks have answers that match with my first method; which seems bizarre.

  • $\begingroup$ Your first step is invalid. Shifting of variables gives the same result only for convergent and logarithmically divergent integrals. Your integral is linearly divergent and hence, if you shift the variables, you need to account for an additional shift in the whole integral. Cf. Schwartz 30.2.2 (p.624). $\endgroup$ Feb 14, 2018 at 16:09

1 Answer 1


The reason you run into problems is you haven't really regularized your loop integral. You're dealing with an ill-defined quantity, and it is natural that you run into contradictions.

The way to go is to regularize first (which in essence defines your object of interest), and then manipulate it. You'll then see that some formal operations you use are justified, and some aren't.

If you use momentum cut-off regularization, in general you can't shift anymore (since integration domain is no longer infinite). Similarly, to perform analytical continuation in space-time dimension you usually employ spherical cooedinates, hence lose manifest translation invariance. Ofcourse the shift invariance will come back upon lifting regularization if you're dealing with a convergent integral to begin with. Not always true otherwise.

Also, your expression is a matrix, the size of which depends on space-time dimension. It is hard to make sense of a matrix with non-integer number of rows and columns. It would be a good idea to work with the final amplitude, which is scalar.


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