$$\int_0^\infty -\arcsin e^{-x}dx$$ Okay so, I thought about somehow transforming it into $\arctan(f(x))$ and then adding the same integral and then somehow use $\arctan(x)+\arctan(\frac{1}{x})=\frac{π}{2}$ but didn't end up anywhere similar to so. I also thought about having some substitution so that I can add a "0 integral", an integral whose value is 0 because it is odd, so that I can somehow try to cancel a pesky factor after combining those 2 integrals? But that didn't work out. Hope it's understandable that I don't write that work because it didn't work and it would be 100 lines long. Is this integral plausible with non-advanced methods? Like Feynman's technique or something.
My attempt $\int_0^\infty -\arcsin e^{-x}dx$
definite integralsimproper-integralsintegration
Related Solutions
A direct proof for $\int_0^x \frac{- x \ln(1-u^2)}{u \sqrt{x^2-u^2}} \, \mathrm{d} u = \arcsin^2(x)$
Let $u=x \sin (\theta)$ \begin{eqnarray*} -\int_0^{\pi/2} \frac{\ln(1-x^2 \sin^2(\theta))}{\sin(\theta)} d \theta \end{eqnarray*} Now expand the logarithms \begin{eqnarray*} \sum_{n=1}^{\infty} \int_0^{\pi/2} \frac{1}{n} x^{2n} \sin^{2n-1}(\theta) d \theta \end{eqnarray*} Now use \begin{eqnarray*} \int_0^{\pi/2} \sin^{2n-1}(\theta) d \theta= \frac{(2n-2)!!}{(2n-1)!!}. \end{eqnarray*} Finally use the result you state in the question \begin{eqnarray*} \frac{1}{2} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} x^{2n}=(\sin^{-1}(x))^2 \end{eqnarray*} and we are done.
Solution 1.
By splitting the integral at $1$ and letting $x\to \frac{1}{x}$ in the second part, we get:$$I=\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx=\int_0^1 \frac{\ln(1+x+x^2)+\ln\left(1+\frac{1}{x}+\frac{1}{x^2}\right)}{1+x^2}dx$$ $$=2\int_0^1 \frac{\ln(1+x+x^2)}{1+x^2}dx-2\int_0^1 \frac{\ln x}{1+x^2}dx$$ Via the substitution $x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt$ and using this, we obtain: $$I=2\int_0^1\frac{\ln\left(\frac{3+t^2}{(1+t)^2}\right)}{1+t^2}dt+2G=2\int_0^1 \frac{\ln(3+t^2)}{1+t^2}dt-4\int_0^1\frac{\ln(1+t)}{1+t^2}+2G$$ The second one is a well known Putnam integral, and for the first one we can try to use Feynman's trick. $$I=2J-\frac{\pi}{2}\ln 2+2G, \quad J=\int_0^1 \frac{\ln(3+x^2)}{1+x^2}dx$$
$$J(a)=\int_0^1 \frac{\ln(2+a(1+x^2))}{1+x^2}dx\Rightarrow J'(a)=\frac1a\int_0^1 \frac{dx}{\frac{a+2}{a}+x^2}dx$$ $$=\frac1a\sqrt{\frac{a}{a+2}}\arctan\left(x\sqrt{\frac{a}{a+2}}\right)\bigg|_0^1=\frac{1}{\sqrt{a(a+2)}}\arctan\left(\sqrt{\frac{a}{a+2}}\right)$$ We are looking to find $J=J(1)$, but we also have: $J(0)=\frac{\pi}{4}\ln 2$ so: $$J=J(1)-J(0)+J(0)=\underbrace{\int_0^1 J'(a)da}_{=K}+\frac{\pi}{4}\ln 2 $$ Now letting $\sqrt{\frac{a+2}{a}}=x\Rightarrow \frac{1}{\sqrt{a(a+2)}}da=-a dx=-\frac{2}{x^2-1}dx\,$ gives us: $$K=\int_0^1 \frac{1}{\sqrt{a(a+2)}}\arctan\left(\sqrt{\frac{a}{a+2}}\right)da=2\int_\sqrt 3^\infty \frac{\arctan \left(\frac{1}{x}\right)}{x^2-1}dx$$ $$=\frac{\pi}{2}\ln(2+\sqrt 3)-2\int_{\sqrt 3}^\infty \frac{\arctan x}{x^2-1}dx $$ $$H=2\int_{\sqrt 3}^\infty \frac{\arctan x}{x^2-1}dx\overset{x=\tan t}=-2\int_\frac{\pi}{3}^\frac{\pi}{2} \frac{t}{\cos(2t)}dt\overset{\large 2t=x+\frac{\pi}{2}}=\int_{\frac{\pi}{6}}^\frac{\pi}{2} \frac{\frac{\pi}{4}+\frac{x}{2}}{\sin x}dx$$ $$=\frac{\pi}{4}\ln\left(\tan\frac{x}{2}\right)\bigg|_\frac{\pi}{6}^\frac{\pi}{2}+\frac12 \int_0^\frac{\pi}{2}\frac{x}{\sin x}dx-\frac12\int_0^\frac{\pi}{6}\frac{x}{\sin x}dx$$ The last two integrals are linked in this post and using their values we get: $$H=\frac{\pi}{4}\ln(2+\sqrt 3)+G+\frac{\pi}{12}\ln(2+\sqrt 3)-\frac23G=\boxed{\frac{\pi}{3}\ln(2+\sqrt 3)+\frac13G}$$ $$\Rightarrow \boxed{K=\frac{\pi}{6}\ln(2+\sqrt 3)-\frac13G}\Rightarrow \boxed{J=\frac{\pi}{6}\ln(2+\sqrt 3)+\frac{\pi}{4}\ln 2-\frac13G}$$ $$\Rightarrow I=\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx=\boxed{\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G}$$
Solution 2.
We can start by considering: $$A=\int_0^\frac{\pi}{2} \ln(2+\sin x)dx,\quad B=\int_0^\frac{\pi}{2}\ln(2-\sin x)dx$$ Like in mrtaurho's approach we have: $$I=\frac{\pi}{2}\ln 2 +A=\frac{\pi}{2}\ln 2+\frac12\left((A+B)+(A-B)\right)\tag 1$$ A solution for $A-B\,$ can be found here. $$A-B=\int_0^\frac{\pi}{2}\ln\left(\frac{2+\sin x}{2-\sin x}\right)dx=-\frac{\pi}{3}\ln(2+\sqrt 3) +\frac{8}{3}G\tag2$$ And for $A+B$ we can directly use this result. $$A+B=\int_0^\frac{\pi}{2} \ln(4-\sin^2 x)=\int_0^\frac{\pi}{2} \ln(4\cos^2x +3\sin^2 x)dx$$$$=\pi \ln 2 +\int_0^\frac{\pi}{2} \ln\left(\cos^2 x+\frac34 \sin^2 x\right)dx=\pi\ln\left(1+\frac{\sqrt 3}{2}\right)\tag3$$ Now plugging $(2)$ and $(3)$ into $(1)$ yields the result.
$$\boxed{I=\frac{\pi}{2}\ln 2+\frac12\left(\pi\ln(2+\sqrt 3)-\pi \ln 2-\frac{\pi}{3}\ln(2+\sqrt 3)+\frac83G\right)=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G}$$
Best Answer
$$I=\int_{0}^{\infty} -\sin^{-1}(e^{-x})dx$$ $$I= -\int_{0}^{\infty} \sin^{-1}(e^{-x})dx$$ Let, $e^{-x}=u \implies dx=\dfrac{-du}{u}$
$$I= -\int_{1}^{0} \frac{\sin^{-1}(u)}{u}(-du)$$ $$I= -\int_{0}^{1} \frac{\sin^{-1}(u)}{u} du$$ Let, $\sin^{-1}(u)=z \implies u=\sin z \implies du=\cos z dz$
$$I= -\int_{0}^{\pi/2} \frac{z}{\sin z}\cos z dz$$ $$I= -\int_{0}^{\pi/2} z\cot z dz$$ By Integration By Parts and then by further evaluating the integral, $$I=-\dfrac{\pi}{2}\ln(2)$$