# 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.

• Mathematica gives $-\pi \ln(2)/2$ Aug 13, 2021 at 22:25
• Why not a simple $t=e^x$ ? Aug 13, 2021 at 22:26
• Maybe you can substitute for one of the terms? Aug 13, 2021 at 22:27
• u-substitution gives $-\int_0^1 \frac{dz}{z} sin ^{-1}(z)$ which is given here: functions.wolfram.com/ElementaryFunctions/ArcSin/21/02/01/0002 Would have to study to obtain that from scratch though. Calculus of residues? Series representation? Feynman's trick? Aug 13, 2021 at 22:31

$$I=\int_{0}^{\infty} -\sin^{-1}(e^{-x})dx$$ $$I= -\int_{0}^{\infty} \sin^{-1}(e^{-x})dx$$ 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, $$I=-\dfrac{\pi}{2}\ln(2)$$

• How do you get the final step by IBP? Aug 14, 2021 at 6:39
• The final step does not follow by IBP. Aug 14, 2021 at 14:32
• How do you evaluate the final integral? It does integrate into a polylogarithm, but shouldn't you show this? It's the hardest part of the problem by far. Aug 14, 2021 at 23:50

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

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.

• Unnecessary complexity to arrive at $F(1)$. Simply substituting $t = e^{-x}$ into the original integral arrives at precisely the same result. Aug 13, 2021 at 22:59