Calculate improper integral $\displaystyle \int_0^1\dfrac{\arcsin(x)}{\sqrt{1-x^2}}\,dx$
We had the following equation to calculate improper integrals (2nd style):
Let f in$\left(a,b\right]$ unbounded, but $\forall \varepsilon >0$ in every subinterval $\left[a+\varepsilon,b\right]$ is bounded, we define:
$\displaystyle \int_a^b f(x)\,dx:=\lim_{\varepsilon\to0^+}\displaystyle \int_{a+\varepsilon}^b f(x)\, dx$
However, I only came up with this solution:
\begin{align}
&\displaystyle \int_0^1 \dfrac{\arcsin(x)}{\sqrt{1-x^2}}\,dx &&\ \mid \ u=\arcsin x \to dx=\sqrt{1-x^2}\, du\quad u=\arcsin0=0 \quad u=\arcsin1=\pi/2\\
&=\displaystyle \int_0^{\frac\pi2}u\,du\\
&= \left[\frac{u^2}{2}\right]^{\frac\pi2}_0\\
&=\frac{\pi^2}{8}
\end{align}
My question is now, which is more accurate (and is this even correct)?
Best Answer
As mentioned in another answer, the "improperness" occurs at $x=1.$ So you should let $0<b<1$ and consider
$$\int_0^b\frac{\arcsin(x)}{\sqrt{1-x^2}}\,dx,$$ then take the limit as $b\to 1^-.$ This will lead to the answer you found.