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$$\int_0^\infty -\arcsin e^{-x}dx$$ 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.

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  • $\begingroup$ Mathematica gives $-\pi \ln(2)/2$ $\endgroup$ Commented Aug 13, 2021 at 22:25
  • $\begingroup$ Why not a simple $t=e^x$ ? $\endgroup$
    – Lucian
    Commented Aug 13, 2021 at 22:26
  • $\begingroup$ Maybe you can substitute for one of the terms? $\endgroup$
    – Alex K
    Commented Aug 13, 2021 at 22:27
  • $\begingroup$ 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? $\endgroup$ Commented Aug 13, 2021 at 22:31

3 Answers 3

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

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    $\begingroup$ How do you get the final step by IBP? $\endgroup$ Commented Aug 14, 2021 at 6:39
  • $\begingroup$ The final step does not follow by IBP. $\endgroup$ Commented Aug 14, 2021 at 14:32
  • $\begingroup$ 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. $\endgroup$ Commented Aug 14, 2021 at 23:50
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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

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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.

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

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