Exercise 15.1.3

Proof. The map $t\mapsto {(1-t^4)}^{-1∕2}$ is continuous on $[0,1[$, thus $t\mapsto {(1-t^4)}^{-1∕2}dt$ is summable on $[1,x]$ for all $x\in [0,1]$.

Since $1-t^4=(1-t)(1+t+t^2+t^3)$, ${(1-t^4)}^{-1∕2}\sim {[4(1-t)]}^{-1∕2}$ in the neighborhood of $1$. The Riemann Criterium shows that ${\int }_0^1{(1-t)}^{-\alpha }dt$ converges if $\alpha <1$, and here $\alpha =1∕2$. Since ${(1-t^4)}^{-1∕2}>0$, this is sufficient to prove that ${\int }_0^1{(1-t^4)}^{-1∕2}dt$ converges.

Since $t\mapsto {(1-t^4)}^{-1∕2}$ is even, the same is true in the neighborhood of $-1$, thus ${\int }_{-1}^0{(1-t^4)}^{-1∕2}dt$ converges. □