Angles, logs and 𝟚

The function is only defined for $x\in(0,1]$ . When $x=0$ the function is undefined but we can evaluate the limit from the right (since negative values are undefined) to check if it has one and stablish if the shape of the curve. After two iterations of L'Hopital's Rule we find the limit is $0$ . We know that $f(1)=0$ so the function starts and ends at $0$ in the interval $x\in (0,1]$ . To find the average value of the curve we need to integrate the function and divide by the length of the interval in the x-axis, since this will give us the height of the rectangle whose area is equal to the area of the curve: $\overline{f(x)}=\frac{1}{1-0}\int_0^1\arcsin x\ln x \hspace{2px}dx=\int_0^1\arcsin x\ln x \hspace{2px}dx$ First we use integration by parts, differentiating the log and integrating $\arcsin x$ . So first we solve the latter: $\int \arcsin x \hspace{2px}dx$ Another integration by parts, integrating 1 and differentiating the inverse sine: $\int \arcsin x \hspace{2px}dx=x\arcsin x-\int \frac{x}{\sqrt{1-x^2}}\hspace{2px}dx$ $\text{Let} \hspace{3px} u=1-x^2$ $du=-2x\hspace{2px}dx$ $\implies\int \frac{x}{\sqrt{1-x^2}}\hspace{2px}dx=-\frac{1}{2}\int \frac{1}{\sqrt{u}}\hspace{2px}du$ $=x\arcsin x +\frac{1}{2}\left(2\sqrt{u}\right)=x\arcsin x +\sqrt{1-x^2}$ Plug this into the previous integration by parts: $\int\arcsin x\ln x \hspace{2px}dx=\ln x\left(x\arcsin x +\sqrt{1-x^2}\right)-\int\frac{x\arcsin x +\sqrt{1-x^2}}{x}\hspace{2px}dx$ $=\ln x\left(x\arcsin x +\sqrt{1-x^2}\right)-\int\arcsin x\hspace{2px}dx -\int\frac{\sqrt{1-x^2}}{x}\hspace{2px}dx$ $=(\ln x-1)\left(x\arcsin x +\sqrt{1-x^2}\right)-\int\frac{\sqrt{1-x^2}}{x}\hspace{2px}dx$ $\text{Let} \hspace{3px} x=\cos \theta$ $dx=-\sin\theta\hspace{2px}d\theta$ $\implies\int\frac{\sqrt{1-x^2}}{x}\hspace{2px}dx=-\int\frac{\sin\theta}{\cos\theta}\cdot\sin\theta\hspace{2px}d\theta$ $=-\int\frac{\sin^2\theta}{\cos\theta}\hspace{2px}d\theta=-\int\frac{1-\cos^2\theta}{\cos\theta}\hspace{2px}d\theta$ $=-\int\sec\theta\hspace{2px}d\theta+\int \cos\theta\hspace{2px}d\theta$ $=-\ln|\sec\theta+\tan\theta|+\sin\theta=-\ln\left|\frac{1}{x}+\frac{\sqrt{1-x^2}}{x}\right|+\sqrt{1-x^2}$ $\implies\int_0^1\arcsin x\ln x \hspace{2px}dx=\left|(\ln x-1)\left(x\arcsin x +\sqrt{1-x^2}\right)-\left(-\ln\left|\frac{1}{x}+\frac{\sqrt{1-x^2}}{x}\right|+\sqrt{1-x^2}\right) \right|_0^1$ Now, we evaluate the integral. When $x=1$ , $F(x)=-\frac{\pi}{2}$ , but when $x=0$ $F(x)$ is undefined, so we take the limit from the right (since negative values are also undefined). To make the limit easier we can rewrite $F(x)$ as follows: $F(x)=\ln \left(\sqrt{1-x^2}+1\right)+\ln \left(x^{x\arcsin x}\right)+\ln\left(x^{\sqrt{1-x^2}-1}\right)-2\sqrt{1-x^2}-x\arcsin x$ $\lim_{x\to 0^+} \left[\ln \left(\sqrt{1-x^2}+1\right)+\ln \left(x^{x\arcsin x}\right)+\ln\left(x^{\sqrt{1-x^2}-1}\right)-2\sqrt{1-x^2}-x\arcsin x\right]$ $=\lim_{x\to 0^+} \left[\ln 2+\ln \left(x^{x\arcsin x}\right)+\ln\left(x^{\sqrt{1-x^2}-1}\right)-2-0\right]$ $=\ln 2-2+\ln \left(\lim_{x\to 0^+}x^{x\arcsin x}\right)+\ln\left(\lim_{x\to 0^+}x^{\sqrt{1-x^2}-1}\right)$ $=\ln 2 -2+\ln\left(e^{\lim_{x\to 0^+}(\ln x)(x\arcsin x)}\right)+\ln\left(e^{\lim_{x\to 0^+}(\ln x)(\sqrt{1-x^2}-1)}\right)$ $\text{After some L'Hopital's Rule shenanigans:}$ $=\ln 2 -2+0+0=\ln2 -2$ $\therefore \left|(\ln x-1)\left(x\arcsin x +\sqrt{1-x^2}\right)-\left(-\ln\left|\frac{1}{x}+\frac{\sqrt{1-x^2}}{x}\right|+\sqrt{1-x^2}\right) \right|_0^1=-\frac{\pi}{2}-(\ln 2-2)$ $\implies\overline{f(x)}=2-\frac{\pi}{2}-\ln 2\approx \boxed{-0.264}$

Here's another solution that avoids some of the difficulty at $x=0$ by choosing a different integration constant.

Assume $f(x)$ is supposed to be a real-valued function with the biggest possible domain, i.e. $f:\:(0;\:1]\rightarrow\mathbb{R}$ . Then we want to calculate $\mu:=\frac{1}{1-0}\int_{0}^1f(x)\:dx=\int_{0}^1f(x)\:dx,\qquad f(x)=\arcsin(x)\ln(x)=\frac{\ln(x)}{\arcsin^{-1}(x)}$ We need to make sure the integral converges at the lower border $x\rightarrow 0^+$ . We use l'Hopital's rule twice to notice $f$ is continuous from above at $x=0$ : $\lim_{x\rightarrow 0^+}f(x) \underset{\text{l'H}}{=}\lim_{x\rightarrow 0^+}-\arcsin^2(x)\sqrt{1-x^2}\cdot \frac{1}{x}\underset{\text{l'H}}{=} \lim_{x\rightarrow 0^+}-\arcsin(x)\cdot\cancel{\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}}=0$ Now we know the integral converges, we want to simplify it with integration by parts. We will need an anti-derivative of $\arcsin(x)$ : $\begin{aligned} \int \arcsin(x)\:dx&=x\arcsin(x) + \sqrt{1-x^2}+C&\left|\text{Choose }C=-1\right.&&&&&(*) \end{aligned}$ We choose the integration constant $C=-1$ to avoid further convergence trouble at $\red{x=0}$ . Using integration by parts, we get $\begin{aligned} \mu&=\left. \left( x\arcsin(x) + \sqrt{1-x^2}-1 \right)\ln(x) \right|_{\red{0}}^1 - \int_0^1\left( x\arcsin(x) +\sqrt{1-x^2}-1 \right)\cdot\frac{1}{x}\:dx\\[.5em] &=0-\red{0}-\int_0^1\arcsin(x) + \frac{ (1-x^2)-1 }{ \sqrt{1-x^2}+1 }\cdot\frac{1}{x}\:dx\underset{(*)}{=}-\frac{\pi}{2}+1+\int_0^1\frac{x\:dx}{1+\sqrt{1-x^2}} \end{aligned}$ To evaluate the lower border, we use l'Hospitals Rule on $x\ln(x)$ : $\lim_{x\rightarrow 0^+}\left(x\arcsin(x) + \sqrt{1-x^2}-1\right)\ln(x)=\lim_{x\rightarrow 0^{+}}\left( \arcsin(x) - \frac{x}{\sqrt{1-x^2}+1} \right)\cdot x\ln(x)\underset{\text{l'H}}{=} \lim_{x\rightarrow 0^+} \left( \arcsin(x) - \frac{x}{\sqrt{1-x^2}+1} \right)\cdot (-x)=0$ We solve the remaining integral by substitution $x=\sin(t),\: dx=\cos(t)\:dt$ : $\begin{aligned} && \int_0^1\frac{x\:dx}{1+\sqrt{1-x^2}}&=\int_0^{\frac{\pi}{2}} \frac{\sin(t)\cos(t)\:dt}{1+\cos(t)}=\int_0^{\frac{\pi}{2}}\sin(t)-\frac{\sin(t)}{1+\cos(t)}\:dt=\left[ -\cos(t) + \ln|1+\cos(t)|\right]_0^{\frac{\pi}{2}}=1-\ln(2) \end{aligned}$
Putting all together, we find $\mu = -\frac{\pi}{2}+2-\ln(2)\approx\boxed{-0 .264}$