Assigning a finite value to a divergent integral 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 :)
 A: 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
$${}_2F_1\left(-\frac{1}{2},\frac{3}{2};\frac{1}{2};z\right)
= \frac{1-2z}{\sqrt{1-z}}.$$
A: $$x = \sin^2(\theta) \implies I = \int_0^{\pi/4} \dfrac{2 \sin(\theta) \cos(\theta)}{\sin^3(\theta) \cos^3(\theta)} d\theta$$
\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 $I = 0$.
