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Can someone help me to regularize the following divergent integral?

$$ \int_0^{1/2}\, \frac{d x}{x^{3/2} (1-x)^{3/2}} $$

Guys, thank you very much for your answers. Thus if I have understood your procedure, the regularized result of this divergent integral (let's do a trivial case) $$ \int_0^\infty{dx} = \lim_{\Lambda\rightarrow \infty} \int_0^\Lambda{dx}=\lim_{\Lambda\rightarrow \infty}\Lambda - 0 \equiv 0 $$ is zero because one simply remove the divergency and the game is over, right?

Well, I would like to have your opinion about this other regularization I have thought of $$ \int_0^\infty{dx} = \lim_{m\rightarrow\infty} \int_0^m{dx} = \lim_{m\rightarrow\infty} 1+\sum_{n=1}^{m-1} 1 = \lim_{m\rightarrow\infty} 1+\sum_{n=1}^{m-1} {1\over n^0} = 1+\zeta(0)=1-{1\over 2}={1\over 2} $$ where I have used the well-known value $\zeta(0)=-1/2$ of the Riemann $\zeta$-function. I was wondering what can be the physical interpretation of such a (naive, I admit) regularization...but maybe there is none and I am just a crazy physicist :)

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What do you mean? –  Gerry Myerson Sep 17 '12 at 12:38
This integral formally diverges at the lower end. However, I know that there are ways to regularize the integral and assign it a finite value (by throwing away a formally infinite quantity, of course). –  Andrea Sep 17 '12 at 12:52
Well, if that's all you want, pick your favorite number between 0 and 1/2 (not including 0), and throw away the integral from zero to your number, and report the finite value you get by integrating from your number to 1/2. If you want more than this, then you have to know what you want, what conditions your regularization has to satisfy. –  Gerry Myerson Sep 17 '12 at 13:05
The first question, and @oen's answer, arose in J. Hadamard's "finite partie" idea, which had a similarly disturbingly ad-hoc appearance, but which he used self-consistently to some advantage. At the same time, I'd agree with Gerry M. that one definitely needs some sense of the "function" of the regularization, since, without constraints, a moderately clever person can arrange to get any answer they want, etc. –  paul garrett Oct 4 '12 at 16:22

2 Answers 2

Here's a fairly natural way to assign the integral a value.

Introduce a cutoff, $$\begin{eqnarray*} I_\epsilon &=& \int_\epsilon^{1/2} \frac{dx}{x^{3/2}(1-x)^{3/2}} \\ &=& \frac{2(1-2\epsilon)}{\sqrt{\epsilon(1-\epsilon)}} \\ &=& \frac{2}{\sqrt{\epsilon}}-3\sqrt{\epsilon} + O(\epsilon^{3/2}). \end{eqnarray*}$$ The simplest regularization is to remove only the divergence, in which case we find the regularized integral is zero: $$I_{\mathrm{reg}} = \lim_{\epsilon\to0}\left(I_\epsilon-\frac{2}{\sqrt{\epsilon}}\right) = 0.$$

Another approach involves analytic continuation of the incomplete beta function, with the same result.

Addendum: Consider the integral representation of the incomplete beta function, $$B_z(a,b) = \int_0^z dx\, x^{a-1}(1-x)^{b-1}.$$ This representation requires $\mathrm{Re}(a) > 0$. (We assume $0<z<1$.) In terms of the hypergeometric function, $$B_z(a,b) = \frac{z^a}{a} {}_2 F_1(a,1-b;a+1;z).$$ We take the right hand side to define the integral, wherever it makes sense to do so. Thus, the integral is $$\begin{eqnarray*} B_{1/2}\left(-\frac{1}{2},-\frac{1}{2}\right) &=& -2\sqrt{2} \ {}_2F_1\left(-\frac{1}{2},\frac{3}{2};\frac{1}{2};\frac{1}{2}\right) \\ &=& 0. \end{eqnarray*}$$ The last equality is not trivial. In fact $${}_2F_1\left(-\frac{1}{2},\frac{3}{2};\frac{1}{2};z\right) = \frac{1-2z}{\sqrt{1-z}}.$$

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Could you give a brief description of how the analytic continuation of the incomplete beta function is carried out? I'm very interested. –  Antonio Vargas Sep 19 '12 at 1:15
@AntonioVargas: Sure, Antonio. I've added something above. –  user26872 Sep 19 '12 at 2:09
Cheers, thanks :) –  Antonio Vargas Sep 19 '12 at 2:23

$$x = \sin^2(\theta) \implies I = \int_0^{\pi/4} \dfrac{2 \sin(\theta) \cos(\theta)}{\sin^3(\theta) \cos^3(\theta)} d\theta$$

\begin{align} I & = 8 \int_0^{\pi/4} \dfrac{d\theta}{4\sin^2(\theta) \cos^2(\theta)} = 8 \int_0^{\pi/4} \dfrac{d \theta}{\sin^2(2 \theta)}\\ & = 8 \int_0^{\pi/4} \text{cosec}^2(2 \theta) d \theta = 4 \int_{0}^{\pi/2} \text{cosec}^2(\phi) d \phi\\ & = -4 \left. \cot(\phi) \right \vert_{0}^{\pi/2} = \lim_{t \to 0} 4 \cot(t) \\ & = 4 \left( \dfrac1t - \dfrac{t}3 - \dfrac{t^3}{45} - \dfrac{2t^5}{945} + \mathcal{O} \left(t^7 \right) \right)\end{align} Removing the pole at $t=0$, and substituting $t=0$ in the rest, we get $I = 0$.

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