Series expansion (likely Maclaurin) of integral

Question

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

Answer 1

If you want to find an expansion of an integral, then start off with the expansion for $\arcsin x$ which is simply$$\arcsin x=\sum\limits_{n\geq0}\binom {2n}n\frac {x^{2n+1}}{4^n(2n+1)}$$And divide the expansion by $x$.$$\frac {\arcsin x}x=\sum\limits_{n\geq0}\binom {2n}n\frac {x^{2n}}{4^n(2n+1)}$$Now integrate term wise to get$$\int\limits_0^{\tfrac 1{\sqrt2}}\mathrm dx\,\frac {\arcsin x}x\color{blue}{=\sum\limits_{n\geq0}\binom {2n}n\frac 1{4^n(2n+1)^2 2^{n+1/2}}}$$

Answer 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 $$I=\sin ^{-1}(x) \log \left(1-e^{2 i \sin ^{-1}(x)}\right)-\frac{1}{2} i \left(\sin ^{-1}(x)^2+\text{Li}_2\left(e^{2 i \sin ^{-1}(x)}\right)\right)$$ making for $$J=\int_0^a \frac{\sin ^{-1}(x)}{x}\,dx$$ $$J=\frac{i \pi ^2}{12}-\frac{1}{2} i \left(\text{Li}_2\left(-2 a^2+2 i \sqrt{1-a^2} a+1\right)+\sin ^{-1}(a) \left(\sin ^{-1}(a)+2 i \log \left(2 a \left(a-i \sqrt{1-a^2}\right)\right)\right)\right)$$ For sure, as you did, we could use the Taylor expansion and get $$J=\sum^{\infty}_{n=0} \frac{ (2 n)! }{4^{n}(2 n+1)^2 (n!)^2}a^{2 n+1}=a+\frac{a^3}{18}+\frac{3 a^5}{200}+\frac{5 a^7}{784}+\frac{35 a^9}{10368}+\frac{63 a^{11}}{30976}+\frac{231 a^{13}}{173056}+O\left(a^{15}\right)$$ 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 $$g(x)=1-\frac{5 x^2}{6}-\frac{11 x^4}{120}-\frac{17 x^6}{560}-\frac{115 x^8}{8064}-\frac{203 x^{10}}{25344}-\frac{735 x^{12}}{146432}-\frac{451 x^{14}}{133120}-\frac{6721 x^{16}}{2785280}+O\left(x^{18}\right)$$ 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 $0.7301811$.