# About $\int_{-\infty}^{0}\text{arcsin}\left(e^x\right)\text{d}x$

I've shown that the integral $$I$$ exists $$I=\int_{-\infty}^{0}\text{arcsin}\left(e^x\right)\text{d}x \approx 1.089$$

Is there a way to find the exact value ? Using $$u=e^x$$ we have $$\text{d}u=u\text{d}x$$ so $$I=\int_{0}^{1}\frac{\text{arcsin}\left(u\right)}{u}\text{d}u$$ But it did not help me.

Applying IBP on your last integral followed by the substitution $$u\mapsto \sin(u)$$ we obtain that

\begin{align*} I=\int_0^1\frac{\arcsin(u)}u\mathrm du&=\underbrace{\left[\log(u)\arcsin(u)\right]_0^1}_{\to 0}-\int_0^1\frac{\log(u)}{\sqrt{1-u^2}}\mathrm du\\ &=-\int_0^{\pi/2}\log(\sin u)\mathrm du \end{align*}

There are various way to evaluate the latter integral. I will use for the sake of simplicity the Fourier Series Expansion of $$\log(\sin u)$$ here. Thus, we further get that

\begin{align*} I=-\int_0^{\pi/2}\log(\sin u)\mathrm du&=-\int_0^{\pi/2}\left[-\log(2)-\sum_{k=1}^\infty \frac{\cos(2ku)}{k}\right]\mathrm du\\ &=\frac\pi2\log(2)+\int_0^{\pi/2}\sum_{k=1}^\infty \frac{\cos(2ku)}k\mathrm du\\ &=\frac\pi2\log(2)+\sum_{k=1}^\infty\frac1k\int_0^{\pi/2}\cos(2ku)\mathrm du\\ &=\frac\pi2\log(2)+\sum_{k=1}^\infty\frac1k\underbrace{\left[\frac{\sin(2ku)}{2k}\right]_0^{\pi/2}}_{=0} \end{align*}

$$\therefore~I~=~\int_0^1\frac{\arcsin(u)}u\mathrm du~=~\frac\pi2\log(2)$$

Note that one could also invoke the Clausen Function aswell as the first derivative of the Beta Function in order to evaluate the logarithmo-trigonometric integral.

Next write $$y=\arcsin u$$ so $$\mathrm{d}u=\cos y\mathrm{d}y$$ and $$I=\int_0^{\pi/2}y\cot y\mathrm{d}y$$. This is a famous integral. One way to compute it is with Feynman's trick, viz. $$I(k)=\int_0^{\pi/2}\frac{\arctan (k\tan y)}{\tan y}\mathrm{d}y=\int_0^k\mathrm{d}k^\prime\int_0^{\pi/2}\frac{\mathrm{d}y}{1+k^{\prime2}\tan^2 y}\\=\int_0^k\mathrm{d}k^\prime\int_0^{\infty}\frac{{d}t}{(1+t^2)(1+k^{\prime2}t^2)}.$$The rest is an exercise in partial fractions, eventually obtaining$$I(k)=\frac{\pi}{2}\int_0^k\frac{\mathrm{d}k^\prime}{1+k^\prime}=\frac{\pi}{2}\ln|1+k|.$$The case $$k=1$$ agrees with mrtaurho's answer.

Use $$\;u:=\sin(t)\;$$ to continue and integrate by parts to get : \begin{align} I&=\int_0^{\large{\frac {\pi}2}}\frac t{\sin(t)}\cos(t)\,dt\\ &=\left.t\,\log(\sin(t))\right|_0^{\large{\frac {\pi}2}}-\int_0^{\large{\frac {\pi}2}}\log(\sin(t))\,dt\\ &=\frac{\pi}2\log(2)\\ \end{align} since the last integral is well known at SE (for example here)