Evaluate $\int_0^1 \frac{\log^3 x}{1+x}\arcsin^2\left(\frac{\sqrt{x}}{2}\right)\mathrm{d}x.$

Question

I'm looking for evaluation of the integral $$\int_0^1 \frac{\log^3 x}{1+x}\arcsin^2\left(\dfrac{\sqrt{x}}{2}\right)\mathrm{d}x.$$ I've tried some trivial substitutions and series expansions so far but it didn't get me anything satisfactory. I'm wondering if there's a closed form for this integral in terms of classical special functions and constants?

Any help would be highly appreciated. Thanks!

Answer 1

Hint

Use a series expansion of $$\frac{\Bigg[\sin ^{-1}\left(\frac{\sqrt{x}}{2}\right)\Bigg]^2}{x+1}=\sum_{n=1}^\infty a_n\,x^n$$ and use (a few integrations by parts) $$\int x^n\,\log^3(x)\,dx =\frac{x^{n+1} \left((n+1)^3 \log ^3(x)-3 (n+1)^2 \log ^2(x)+6 (n+1) \log (x)-6\right)}{(n+1)^4}$$ $$\int_0^1 x^n\,\log^3(x)\,dx =-\frac{6}{(n+1)^4}$$

If you use an expansion to $O\left(x^7\right)$, the decimal representation of the result is $-0.08048$ while numerical integration gives $-0.08069$.

Edit

Much better would be to use $$\Bigg[\sin ^{-1}\left(\frac{\sqrt{x}}{2}\right)\Bigg]^2=\sum_{n=0}^\infty \frac{x^{n+1}}{4 (n+1)^2 \binom{2 n+1}{n}}$$ and $$I_n=\int_0^1 \frac {x^{n+1}}{x+1}\,\log^3(x)\,dx =-\frac{6}{(n+1)^4}+\frac{1}{16} \left(\psi ^{(3)}\left(\frac{n+1}{2}\right)-\psi ^{(3)}\left(\frac{n+2}{2}\right)\right)$$

We know the first part

$$S_1=-\frac{3}{2}\sum_{n=0}^\infty \frac{1}{ (n+1)^6 \binom{2 n+1}{n}}=-\frac{3}{2} \, _7F_6\left(1,1,1,1,1,1,1;\frac{3}{2},2,2,2,2,2;\frac{1}{4}\right)$$ which, numerically is $$S_1=-1.508029564434807859364032\cdots$$

For the second part, if we let $$a_n=-\frac{\psi ^{(3)}\left(\frac{n+2}{2}\right)-\psi ^{(3)}\left(\frac{n+1}{2}\right)}{64 (n+1)^2 \binom{2 n+1}{n}}$$ $$\frac{a_{n+1}}{a_n}=\frac{1}{4}-\frac{11}{8 n}+\frac{85}{16 n^2}-\frac{511}{32 n^3}+O\left(\frac{1}{n^4}\right)$$

$$a_n=\frac{3 \sqrt{\pi } }{2^{2 n+3}\,n^{11/2}}\Bigg[1-\frac{27}{8 n}+\frac{553}{128 n^2}+\frac{191}{1024 n^3}+O\left(\frac{1}{n^4}\right) \Bigg]$$

All the above show that the summation will converge very fast. $$\left( \begin{array}{ccc} p & \sum_{n=0}^p a_n & S_1+\sum_{n=0}^p a_n \\ 0 & 1.42054924424587 & -0.08748032019 \\ 1 & 1.42717014055871 & -0.08085942388 \\ 2 & 1.42732902105033 & -0.08070054338 \\ 3 & 1.42733655546868 & -0.08069300897 \\ 4 & 1.42733707613558 & -0.08069248830 \\ 5 & 1.42733712182456 & -0.08069244261 \\ 6 & 1.42733712655208 & -0.08069243788 \\ 7 & 1.42733712710382 & -0.08069243733 \\ 8 & 1.42733712717440 & -0.08069243726 \\ 9 & 1.42733712718411 & -0.08069243725 \\ 10 & 1.42733712718552 & -0.08069243725 \end{array} \right)$$

Since $a_n$ has a simple upper bound, if we want to know where to stop the summation, we need to solve $$\frac{3 \sqrt{\pi } }{2^{2 n+3}\,n^{11/2}} \leq \epsilon$$ which gives $$n \geq \frac{11}{4 \log (2)}\,\,W\left(\frac{2 \log (2)}{11}\left(\frac{288 \pi }{\epsilon^2}\right)^{\frac 1 {11}}\right)$$ where $W(.)$ is Lambert function.

If $\epsilon=10^{-k}$, for $k \geq 6$, a good approximation is $$n \sim 4.26567+1.68146 k-5.67925 \log (k)$$

Update

Using $n$ terms in the initial expansion gives as a result $$A_n=-\frac {a_n}{b_n}+\frac {c_n}{d_n}\pi^4$$

$A_{100}$ is

$$-\frac{944625310296822422725128343436833995663832190235528173269272430121035643 }{676428576453732757696907435157256036402189073050626346524672000000000000}$$ $$+\frac{499006958747810323870968849569}{36941729688398957008812825600000}\pi^4$$

New elements

The appearance of $\pi^4$ made me thinking about another approach which could be interesting for the generalization

$$I_m=\int_0^1\frac{\Bigg[\sin ^{-1}\left(\frac{\sqrt{x}}{2}\right)\Bigg]^2}{x+1} \log^m(x)\,dx$$ since $$\int_0^1 \frac{x^{n+1} }{x+1}\log ^m(x) \,dx=(-1)^m \frac{m!}{2^{m+1}} \left(\zeta \left(m+1,\frac{n+2}{2}\right)-\zeta \left(m+1,\frac{n+3}{2}\right)\right)$$ which gives $$I_m=(-1)^m \frac{m!}{2^{m+3}}\sum_{n=0}^\infty \frac{\zeta \left(m+1,\frac{n+2}{2}\right)-\zeta \left(m+1,\frac{n+3}{2}\right)}{(n+1)^2 \binom{2 n+1}{n}}$$

$$I_m=(-1)^m \sqrt \pi \frac {m!}{2^{m+4}}\sum_{n=0}^\infty \frac{\Gamma (n+1)}{2^{2n}\,(n+1) \Gamma \left(n+\frac{3}{2}\right)}\left(\zeta \left(m+1,\frac{n}{2}+1\right)-\zeta \left(m+1,\frac{n+3}{2}\right)\right)$$

$$I_{2p}=\sqrt \pi \frac {(2p)!}{2^{2(p+2)}}\sum_{n=0}^\infty \frac{\Gamma (n+1)}{2^{2n}\,(n+1) \Gamma \left(n+\frac{3}{2}\right)}\left(\zeta \left(2p+1,\frac{n}{2}+1\right)-\zeta \left(2p+1,\frac{n+3}{2}\right)\right) $$ $$I_{2p+1}=-\frac {\sqrt \pi }{2^{2p+5} }\sum_{n=0}^\infty \frac{\Gamma (n+1)}{2^{2n}\,(n+1) \Gamma \left(n+\frac{3}{2}\right)}\left(\psi ^{(2 p+1)}\left(\frac{n+2}{2}\right)-\psi ^{(2 p+1)}\left(\frac{n+3}{2}\right)\right)$$

Answer 2

Just a beginning.

$$\arcsin^2\left(\frac{\sqrt{x}}{2}\right) = \frac{1}{2}\sum_{n\geq 1}\frac{x^n}{n^2\binom{2n}{n}} $$

and

$$ \int_{0}^{1}\frac{x^n \log^3(x)}{1+x}\,dx = \sum_{m\geq 0}(-1)^m\int_{0}^{1} x^{n+m}\log^3(x)\,dx = -6\sum_{m\geq 0}\frac{(-1)^m}{(m+n+1)^4} $$ so our integral equals $$ 3\sum_{n\geq 1}\frac{(-1)^n}{n^2\binom{2n}{n}}\sum_{m > n}\frac{(-1)^m}{(m+1)^4}=-\frac{7\pi^4}{120}\log^2\left(\frac{1+\sqrt{5}}{2}\right)-3\sum_{n\geq 1}\frac{(-1)^n}{n^2\binom{2n}{n}}\sum_{m=0}^{n}\frac{(-1)^m}{(m+1)^4} $$ which should be related to Euler sums with weigth $6$.

Answer 3

If to take a closer look at the integrand then it turns out that the main line segment that contributes to the integral is concentrated in a small zone near zero.

Extreme point of the integrand lies at $x=0.044$ and after that the integrand quickly approaches zero.

That means we can try to compute (approximately) the integral using the first few terms of the Taylor expansion of $\arcsin^2(x)$

$$\arcsin^2(x)=x^2+\frac{x^4}{3}+...$$

We get

$$I_{approx}=\int_{0}^{1}\frac{\frac{x}{4}+\frac{x^2}{48}}{1+x}\log^3(x)\,dx =\frac{77\pi^4-7965}{5760}$$

The approximation error is about $0.00005$