Checking convergence of an improper integral. Can anyone tell me why the following improper integral converges
$$\int_{-\infty}^0((1-x)^{\frac{1}{4}}-(-x)^\frac{1}{4})^2dx?$$
Thanks!
 A: It does not have a direct relation to the problem (as @Ricardo Cavalcanti mentioned, integral converges), but it would be interesting to evaluate the integral.
$$I=\int_{-\infty}^0((1-x)^{\frac{1}{4}}-(-x)^\frac{1}{4})^2dx=\int_0^{\infty}((1+x)^{\frac{1}{4}}-(x)^\frac{1}{4})^2dx$$
Making change $t=\frac{1}{1+x}$
$$=\int_0^{1}\bigl(t^{-\frac{1}{4}}-(1-t)^\frac{1}{4}t^{-\frac{1}{4}}\bigr)^2\frac{dt}{t^2}=\int_0^{1}\bigl(1-(1-t)^\frac{1}{4}\bigr)^2\frac{dt}{t^\frac{5}{2}}$$
Integrating by part
$$=-\frac{2}{3}\bigl(1-(1-t)^\frac{1}{4}\bigr)^2\frac{1}{t^\frac{3}{2}}|_0^1+\frac{1}{3}\int_0^{1}\bigl(1-(1-t)^\frac{1}{4}\bigr)(1-t)^{-\frac{3}{4}}\frac{dt}{t^\frac{3}{2}}$$ $$=-\frac{2}{3}+\frac{1}{3}\int_0^{1}\Bigl(t^{-\frac{3}{2}}(1-t)^{-\frac{3}{4}}-t^{-\frac{3}{2}}(1-t)^{-\frac{1}{2}}\Bigr)dt$$
We see again that the integral converges - the expression in parentheses of the integrand gives additional power of $t$ at $t\to0$.
Now we want to evaluate the integral $I(\beta,\alpha,\alpha')=\int_0^1s^{\beta-1}\left((1-s)^{\alpha-1}-(1-s)^{\alpha'-1}\right)ds$, where $\beta\in(-1;0)$ and $\alpha, \alpha'\in(0,1)$.
We introduce the analytical continuation of Beta-function ($B(\beta,\alpha)=\int_0^1s^{\beta-1}(1-s)^{\alpha-1}ds$, if $\beta, \alpha >0\,$)
for negative $\beta\in(-1,0)$:
$$B(\beta,\alpha)=-\frac{1}{(\exp(2\pi{i}\alpha)-1)(\exp(2\pi{i}\beta)-1)}\oint_Ps^{\beta-1}(1-s)^{\alpha-1}ds$$
where $P$ is Pochhammer contour in the complex plane.
It can be shown that for $\beta\in(-1;0)$ and $\alpha>0$ $B(\beta,\alpha)=\lim_{r\to0}(\int_r^1s^{\beta-1}(1-s)^{\alpha-1}ds+\frac{r^\beta}{\beta})$,
so $$I(\beta,\alpha,\alpha')=\int_0^1s^{\beta-1}\left((1-s)^{\alpha-1}-(1-s)^{\alpha'-1}\right)ds$$$$=\lim_{r\to0}\int_r^1s^{\beta-1}\left((1-s)^{\alpha-1}-(1-s)^{\alpha'-1}\right)ds=\lim_{r\to0}\left(B(\beta,\alpha)-\frac{r^\beta}{\beta}-B(\beta,\alpha')+\frac{r^\beta}{\beta}\right)$$ $$I(\beta,\alpha,\alpha')=B(\beta,\alpha)-B(\beta,\alpha')$$
It can also be proved that analytically continued Beta-function is expressed in the usual way through Gamma-function: $B(\beta,\alpha)=\frac{\Gamma(\beta)\Gamma(\alpha)}{\Gamma(\beta+\alpha)}$. This expression is valid for all complex $\alpha, \beta\in{C}$
After all we get the final result:
$$I(\beta,\alpha,\alpha')=-\frac{2}{3}+\frac{1}{3}\biggl(B(-\frac{1}{2},\frac{1}{4})-B(-\frac{1}{2},\frac{1}{2})\biggr)=-\frac{2}{3}+\frac{1}{3}\,\Gamma(-\frac{1}{2})\left(\frac{\Gamma(\frac{1}{4})}{\Gamma(-\frac{1}{4})}-\frac{\Gamma(\frac{1}{2})}{\Gamma(0)}\right)$$
$$I=-\frac{2}{3}+\frac{\Gamma^2(\frac{1}{4})}{6\sqrt{2\pi}}$$
