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I came across (coincidentally) two integral evaluations, which seem to agree according to numerical tests. It did not seem easy to convert one into the other.

QUESTION. Is this true? $$\int_0^1\left(\frac{\arcsin x}x\right)^3dx =\frac34\pi\int_0^1\left(2\,\text{arctanh}\, x +\frac{\log(1-x^2)}x\right)dx.$$

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    $\begingroup$ Maple verifies it: L:= int((arcsin(x)/x)^3,x=0..1); $$ L:= -\frac{\pi^3}{16}+\frac{3\pi\,\ln \left( 2 \right)}{2} $$ R:= 3/4*Pi*int(2*arctanh(x)+log(1-x^2)/x,x=0..1); $$ R := \frac{3 \pi \left(-\frac{\pi^2}{12}+2 \ln(2)\right)}{4} $$ normal(L-R); $$ 0$$ $\endgroup$
    – Robert Israel
    Commented Sep 3, 2019 at 3:35
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    $\begingroup$ The question, then, is whether the initial two integrals are equal simply because they are both equal to a third thing, or whether there is a direct transformation. $\endgroup$
    – MyNinthAccount
    Commented Sep 3, 2019 at 4:27
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    $\begingroup$ @MyNinthAccount : The OP's question actually is "Is this true?" Also, I think that, formally speaking, there is no such thing as "a direct transformation". However, there may be a simpler "direct" proof of $A=B$ than an "indirect" proof of $A=C\ \&\ C=B$. $\endgroup$
    – Iosif Pinelis
    Commented Sep 3, 2019 at 12:10
  • $\begingroup$ @IosifPinelis: that'd be my wish. $\endgroup$
    – T. Amdeberhan
    Commented Sep 4, 2019 at 2:47

1 Answer 1

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The proposed equality is true.

Details: To find $$l:=\int_0^1\left(\frac{\arcsin x}x\right)^3\,dx =-\frac{1}{16} \pi \left(\pi ^2-24 \ln2\right), $$ make the substitution $t=\arcsin x$ and repeatedly integrate by parts to kill the powers of $t$ and reduce this integral to $$\int_0^{\pi/2}\ln\sin t\,dt=-\frac\pi2\,\ln2, $$ by formula 4.225.3, page 531, of Gradshteyn--Ryzhik.

To find $$r_1:=\int_0^1 2\,\text{arctanh}\, x\,dx =\ln4, $$ integrate by parts to find an antiderivative of $\text{arctanh}$. Alternatively, expand $\text{arctanh}\, x$ into the Maclaurin series (using $\text{arctanh}'x=\frac1{1-u}=1+u+u^2+\dots$ with $u=x^2$) and integrate termwise, to get $$r_1/2=\sum_1^\infty\frac1{(2j-1)2j} =\sum_1^\infty\Big(\frac1{2j-1}-\frac1{2j}\Big) =\sum_1^\infty\frac{(-1)^{k-1}}k=\ln2. $$

To find $$r_2:=\int_0^1 \frac{\ln(1-x^2)}x\,dx =-\frac12\,\sum_1^\infty\frac1{j^2}=-\frac{\pi^2}{12}, $$ expand $\ln(1-x^2)$ into the Maclaurin series (by using the Maclaurin series for $\ln(1-u)$) and integrate termwise. Now one can see that $l=\frac{3\pi}4\,(r_1+r_2)$, that is, the equality holds.

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    $\begingroup$ You do both RHS integrals termwise in expansions, and reduce the LHS to $\smash{\int\ln\sin}$. As Euler also did the latter termwise in a (“Fourier”) expansion, I wonder (without having checked anything) if maybe the OP’s sought transformation could also be seen “termwise”? Alternatively, loc. cit. may also suggest looking into deformation of appropriate contour integrals. $\endgroup$
    – Francois Ziegler
    Commented Sep 3, 2019 at 6:14
  • $\begingroup$ @FrancoisZiegler : Thank you for the nice "Euler--Fourier" reference. $\endgroup$
    – Iosif Pinelis
    Commented Sep 3, 2019 at 11:59
  • $\begingroup$ Good resolution, thanks. $\endgroup$
    – T. Amdeberhan
    Commented Sep 4, 2019 at 2:47

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