Cfrgtkky

As someone who is trained formally in physics, and not mathematics, I have become rusty in series expansions of special integrals (and/or) identities that exist regarding integrals of inverse trigonometric functions (with special bound). In the context of some of the research I’ve done over the last couple weeks regarding a program, an integral (that I will list below) has repeatedly shown up, with no physical motivation (in the context of verlinde algebras in two dimensional conformal field theories, if y’all are wondering). I reckon a series expansion / identity may make the entire thing make a little more sense!

The integral is:$$int_0^{frac{1}{sqrt{2}}} frac{arcsin x}{x} , dx.$$ I had a similar problem earlier in the program that I was able to evaluate according to some of the Ramanujan identities and found some very interesting results.

I hope the format is not too difficult to understand, and that a more math savvy person knows of any special series expansions / identities regarding the integral.

Thank you all

Here is a way to obtain that closed form. Denote: $$I=int_0^frac{1}{sqrt 2} frac{arcsin x}{x} dxoverset{largearcsin x=t}=int_0^frac{pi}{4}frac{t}{sin t}cos t dt=int_0^frac{pi}{4}t (ln(sin t))'dt$$
$$overset{IBP}=tln(sin t)bigg|_0^frac{pi}{4}-int_0^frac{pi}{4}ln(sin t)dt$$
Note that the last integral appeared here on MSE often, for example here.
$$Rightarrow I=frac{pi}{4}lnleft(frac{1}{sqrt 2}right)+frac12left(G+fracpi2ln2right)=frac{pi}{8}ln2 +frac{G}{2}$$

If you want to find an expansion of an integral, then start off with the expansion for $arcsin x$ which is simply$$arcsin x=sumlimits_{ngeq0}binom {2n}nfrac {x^{2n+1}}{4^n(2n+1)}$$And divide the expansion by $x$.$$frac {arcsin x}x=sumlimits_{ngeq0}binom {2n}nfrac {x^{2n}}{4^n(2n+1)}$$Now integrate term wise to get$$intlimits_0^{tfrac 1{sqrt2}}mathrm dx,frac {arcsin x}xcolor{blue}{=sumlimits_{ngeq0}binom {2n}nfrac 1{4^n(2n+1)^2 2^{n+1/2}}}$$

Assuming that the upper bound is not fixed, let us consider
$$I=int frac{sin ^{-1}(x)}{x},dx$$ As given by a CAS, the result is not the most pleasant
$$I=sin ^{-1}(x) log left(1-e^{2 i sin ^{-1}(x)}right)-frac{1}{2} i left(sin
^{-1}(x)^2+text{Li}_2left(e^{2 i sin ^{-1}(x)}right)right)$$ making for
$$J=int_0^a frac{sin ^{-1}(x)}{x},dx$$ $$J=frac{i pi ^2}{12}-frac{1}{2} i left(text{Li}_2left(-2 a^2+2 i sqrt{1-a^2}
a+1right)+sin ^{-1}(a) left(sin ^{-1}(a)+2 i log left(2 a left(a-i
sqrt{1-a^2}right)right)right)right)$$ For sure, as you did, we could use the Taylor expansion and get
$$J=sum^{infty}_{n=0} frac{ (2 n)! }{4^{n}(2 n+1)^2 (n!)^2}a^{2 n+1}=a+frac{a^3}{18}+frac{3 a^5}{200}+frac{5 a^7}{784}+frac{35 a^9}{10368}+frac{63
a^{11}}{30976}+frac{231 a^{13}}{173056}+Oleft(a^{15}right)$$ which is very quickly convergent. For example, using $a=frac 1 {sqrt{2}}$, this would give $frac{137313678493039}{132975953510400 sqrt{2}}approx 0.730173$ while the exact result Zacky gave is $approx 0.730181$. More terms would made the results more accurate.

Instead of Taylor series, I would prefer to consider that function
$$g(x)=(1-x^2)frac{sin ^{-1}(x)}{x}$$ is a pretty nice function the series expansion of it being
$$g(x)=1-frac{5 x^2}{6}-frac{11 x^4}{120}-frac{17 x^6}{560}-frac{115
x^8}{8064}-frac{203 x^{10}}{25344}-frac{735 x^{12}}{146432}-frac{451
x^{14}}{133120}-frac{6721 x^{16}}{2785280}+Oleft(x^{18}right)$$ As a result, we face integrals
$$K_n=int_0^a frac {x^{2n+1}}{1-x^2}=frac{1}{2} B_{a^2}(n+1,0)$$ where appears the incomplete beta function.

Limited to the above truncation, $a=frac 1 {sqrt{2}}$, this would give $0.7301814$ for an exact value equal to $0.7301811$.