# (Zeta function)Regularization of a divergent integral

Consider the following integral:

$$I(n)=\int_{0}^1dx\; x^{-(2n-2)} (1-x)^{n-1}$$, where $n \in \mathbb{N}$

The integral can be looked upon as a beta integral and hence can be expressed in terms of Gamma functions like so:

$$I(n)=\frac{\Gamma(3-2n) \Gamma(n)}{\Gamma(3-n)}$$

Now, for $n=1$ the integral is absolutely convergent and evaluates to 1.

But, for $n=2$ it is clearly divergent because of the Gamma function in the numerator.

The interesting situation is for $n \geq 3$. In this case both the numerator and denominator blow up but there is a limit one can take:

$$\lim_{x \to n } I(x)=\frac{(-1)^n}{2} \times \frac{(n-3)!(n-1)!}{(2n-3)!}$$

The above result can be shown by using the residue of the Gamma function at negative integers.

So, does it mean that $I(n)$ is conditionally convergent for $n \geq 3$? Also for $n$ odd like $1,3$ the result is negative even though the integrand is non-negative in the domain of integration!

In fact, for $n=3$, $$\lim_{n \to 3} \frac{\Gamma(3-2n)}{\Gamma(3-n)}=-\frac{1}{12}$$

Is $$\frac{\Gamma(-3)}{\Gamma(0)}\overset{?}{=} 1+2+3+\ldots$$

Edit 1:

I am very sorry. I had missed a minus sign in my integral. The post is corrected.

Edit 2:

This integral is clearly not convergent for $n \geq \frac{3}{2}$ as has been rightly pointed out in one of the answers, in any traditional sense like absolute or even conditional. But there seems to be some notion of the integral version of Ramanujan summation or zeta function regularization that may assign this divergent integral some finite answer. This particular integral actually came about in a quantum field theory calculation and is therefore well-motivated in some sense. It would be great if someone could throw some light on this aspect. I will also change the title of the question from "Is this integral convergent(conditionally)?" to reflect this idea.

• I don't know at the moment where the flaw is, but there have to be one. Because of the monotonicity of the integral, when it is convergent, then it has to be positive. Especially for $n=3$ you get $\int_0^1 x^4(1-x)^2 dx = \int_0^1 x^6-2x^5+x^4 dx = \frac17-\frac13+\frac15=\frac1{105}>0$. – Mundron Schmidt Jun 24 '18 at 6:00
• Isnt it simply $B(2n-1, n)$? so it converges when $2n-1>0 & n>0$ ? – markovchain Jun 24 '18 at 6:02
• @markovchain But what about the negative result? – Subho Jun 24 '18 at 6:03
• I think your $I(n)$ expression in terms of gamma fn is worng as it clearly converges for $n=2$ – markovchain Jun 24 '18 at 6:07
• I don't see how to frame it as a zeta regularisation, but it's a regularisation by analytic continuation. For $0 < n < 3/2$, the integral is obviously $\mathrm{B}(3-2n,n)$, and since that is meromorphic on the whole plane, there's your continuation. – Daniel Fischer Jun 24 '18 at 9:45

## 1 Answer

Consider that $$\int_0^{1/2}x^\alpha~dx=\begin{cases}\frac1{\alpha+1}\left(\frac12\right)^{\alpha+1} & \alpha> -1\\\infty & \alpha\leq-1\end{cases}.$$ In your case $-(2n-2)\leq-1$ holds for $n\geq \frac32$. Hence, for $n\geq \frac32$ you get $$\int_0^{1/2}x^{-(2n-2)}(1-x)^{n-1}dx\geq\left(\frac12\right)^{n-1}\int_0^{1/2}x^{-(2n-2)}~dx = \infty.$$ Therefore, your integral doesn't converge for all $n\geq \frac32$.

The same way, you can see that the integral doesn't converge for $n\leq 0$ due to the $(1-x)^{n-1}$ term.

But for $n\in\left(0,\frac32\right)$ it converges to a positive value, because of the monotonicity of the integral! This can be seen through \begin{align} \int_0^{1/2}x^{-(2n-2)}(1-x)^{n-1}~dx\leq\max\left\{1,\left(\frac12\right)^{n-1}\right\}\int_0^{1/2}x^{-(2n-2)}~dx \\=\frac1{3-2n}\max\left\{1,\left(\frac12\right)^{n-1}\right\}\left(\frac12\right)^{3-2n}<\infty \end{align} and \begin{align} \int_{1/2}^1x^{-(2n-2)}(1-x)^{n-1}~dx\leq\max\left\{1,\left(\frac12\right)^{2-2n}\right\}\int_0^{1/2}(1-x)^{n-1}~dx \\=\frac1{n}\max\left\{1,\left(\frac12\right)^{2-2n}\right\}\left(\frac12\right)^{n}<\infty. \end{align}

• I think it should be $\frac{3}{2}$ and not $\frac{1}{2}$ by your reasoning. – Subho Jun 24 '18 at 6:26
• You are right. I fixed it. – Mundron Schmidt Jun 24 '18 at 6:29