My attempt $\int_0^\infty -\arcsin e^{-x}dx$ $$\int_0^\infty -\arcsin e^{-x}dx$$ 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.
 A: Let $$J = -\int_0^\infty \sin^{-1}\big(e^{-x})dx$$
Set $z=e^{-x}$ for
$$ J = - \int_0^1 \frac{\sin^{-1}(z)}{z}dz. $$
Substitute $\sin(u) = z$ for
$$ J = -\int_0^{\pi/2} u \cot(u) du.$$
Integrate by parts for
$$ J = -\int_0^{\pi/2} du \ln(\sin(u)). $$
This equivalent integral is explained here, providing
$$J = -\frac{\pi}{2} \ln(2).$$
Tldr, set $u = \sin^{-1}(e^{-x})$, integrate by parts, go to link
A: $$I=\int_{0}^{\infty} -\sin^{-1}(e^{-x})dx$$
$$I= -\int_{0}^{\infty} \sin^{-1}(e^{-x})dx$$
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,
$$I=-\dfrac{\pi}{2}\ln(2)$$
A: Consider
$$F(s)=\int_0^\infty\arcsin(se^{-x})\mathrm dx$$
We can see that
$$F'(s)=\int_0^\infty\frac{e^{-x}}{\sqrt{1-s^2e^{-2x}}}\mathrm dx$$
We now use a substitution $z=se^{-x},\mathrm dz=-e^{-x}\mathrm dx$ to write this as
$$F'(s)=\frac{1}{s}\int_0^s\frac{\mathrm dz}{\sqrt{1-z^2}}=\frac{\arcsin(s)}{s}$$
One can clearly see that $F(0)=0$ hence what we want is $F(1)$:
$$F(1)=\int_0^1\frac{\arcsin(t)}{t}\mathrm dt$$
We can use the Taylor series of inverse sine:
$$\arcsin(x)=\sum_{n=0}^\infty \frac{(2n)!}{(2n+1)4^nn!^2}x^{2n+1}$$
To write this as an infinite sum
$$F(1)=\sum_{n=0}^\infty \frac{(2n)!}{(2n+1)^24^nn!^2}$$
Though numerically this sum looks similar to the proposed $\frac{\pi}{2}\log 2$, I don't yet have proof of this. Perhaps others can offer some insight on this.
