I'm having some difficulties doing this improper integral:
$$\int_{0}^{1}\frac{1}{\sqrt{1-x^4}}\,dx$$
We have the conflictive point at $1$, but I don't know how to do this integral. I know how to solve this one:
$$\int_{0}^{1}\frac{1}{\sqrt{1-x^2}}\,dx$$
but I don't know how to (or if it's possible) extrapolate this information to the other integral.
Thank you.
Applying the substitution $u=x^4$ we get $x=u^{1/4}$ and $4x^3dx=4u^{3/4}dx=du$ and with some algebra we arrive at:
$$\frac{1}{4}\int_{0}^{1} u^{-3/4}(1-u)^{-1/2} du$$
$$=\frac{1}{4}\int_{0}^{1} u^{\frac{1}{4}-1}(1-u)^{\frac{1}{2}-1} du$$
Which can be evaluated recognizing it is a form of the beta function, and applying the beta-gamma relationship:
$$\frac{1}{4}B(\frac{1}{4},\frac{1}{2})$$
$$=\frac{\frac{1}{4}\Gamma(\frac{1}{4})\Gamma(\frac{1}{2})}{\Gamma(\frac{1}{2}+\frac{1}{4})}$$
$$=\frac{\Gamma(\frac{5}{4})\sqrt{\pi}}{\Gamma(\frac{3}{4})}$$