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I would like to prove that the following integral is logarithmically divergent.

$$\int d^{4}k \frac{k^{4}}{(k^{2}-a)((k-x)^{2}-a)((k-y)^{2}-a)((k-z)^{2}-a)}$$

This is 'obvious' because the power of $k$ in the numerator is $4$, but the highest power of $k$ in the denominator is $8$.

However, it is the highest power of $k$ in the denominator that is $8$. There are other terms in $k$ in the denominator of the form $k^7$, $k^6$, etc.

I was wanting a more formal proof that the integral is logarithmically divergent.

nightmarish
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2 Answers2

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This integral is logarithmically divergent for large $k$. For small $|k|$ your integral should behave regularly. That is given the values of $a,x,y,z$ it should be not difficult to show that $$\left|\int_{|k|\leq 1} d^{4}k \frac{k^{4}}{(k^{2}-a)((k-x)^{2}-a)((k-y)^{2}-a)((k-z)^{2}-a)}\right| \leq M$$ with $M$ some constant (not diverging).

For large $k$ you can approximate the integrand as $1/k^4$ with correction terms that are $O(k^{-3}).$ So introducing a ``cutoff'' $k^*$ for the integral such that we only integrate over the ball $|k| \leq k^*$, your integral is given by $$\left|\int_{1\leq |k|\leq k^*} d^{4}k \frac{k^{4}}{(k^{2}-a)((k-x)^{2}-a)((k-y)^{2}-a)((k-z)^{2}-a)} \right| \leq \int_{1\leq |k|\leq k^*} \frac{d^4 k}{k^4} + C \underbrace{\left|\int_{1\leq |k|\leq k^*}\frac{d^4 k}{k^3}\right|}_{\leq N}$$ with $N$ another constant (independent of $k^*$).

So in total, we have that (for $k^*\to \infty$) $$ \int_{|k|\leq k^*} d^{4}k \frac{k^{4}}{(k^{2}-a)((k-x)^{2}-a)((k-y)^{2}-a)((k-z)^{2}-a)} = \int_{1\leq |k|\leq k^*} \frac{d^4 k}{k^4} + O(1) = S_3 \log (k^*) +O(1)$$ with $S_3= 2\pi^2$ the surface of the 3-sphere.

Fabian
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  • Does your analysis hold true if $k$ is not a Euclidean $4$-vector, but is rather a Lorentzian $4$-vector with metric signature $(+,-,-,-)$ so that $k^{\mu}= (k^{0}, k^{1}, k^{2}, k^{3})$, $k_{\mu}= (k^{0}, -k^{1}, -k^{2}, -k^{3})$ and $k^{2} = (k^{0})^{2} - (k^{1})^{2} - (k^{2})^{2} - (k^{3})^{2}$? – nightmarish Feb 19 '17 at 18:29
  • @nightmarish: yes it does (I was thinking something like this appears, as $k^2-a$ should not have a pole, but you should have mentioned this important fact when writing your question). In this case, you should first perform a Wick rotation $k^0 \to i k^0$. This is possible when there is a proper $\epsilon$ factor included in terms like $k^2-a$ that will guarantee causality. – Fabian Feb 19 '17 at 18:57
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It's been awhile since I've calculated these beastly integrals in QFT but a lot of them can be handled using the introduction of Feynman parameters. The basic version for two factors in the denominator is

$$ \dfrac{1}{AB} = \int_0^1 \int_0^1 dx dy \delta(x+y-1) \dfrac{1}{[xA+yB]^2}$$

For $n$ factors this generalizes to

$$ \dfrac{1}{A_1 A_2 \cdots A_n} = \int_0^1 \int_0^1 \cdots \int_0^1 dx_1 dx_2 \cdots dx_n \delta(\sum_i^{n}x_i-1) \dfrac{(n-1)!}{[x_1A_1+x_2A_2 + \cdots x_nA_n]^2}$$

BobaFret
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  • As $A_{1}=(k^{2}-a)$, $A_{2}=((k-x)^{2}-a)$, $A_{3}=((k-y)^{2}-a)$, $A_{4}=((k-z)^{2}-a)$, there are going to be terms in a linear power of $k$ in the denominator. More specifically, the denominator becomes $[k^{2}(x_{1}+x_{2}+x_{3}+x_{4})-2k(xx_{2}+yx_{3}+zx_{4})-a(x_{1}+x_{2}+x_{3}+x_{4})-x_{2}x^{2}-x_{3}y^{2}-x_{4}z^{2}]^{2}$. – nightmarish Feb 19 '17 at 17:59