Justify an approximation of the first few decimals of $\int_1^\infty(\int_0^1\frac{dx}{-1+x^3+y^2})dy$

Question

This morning I wondered about how to calculate an approximation (I think thus that our integral has no a known indefinite integral, at least for some of the terms in the integrand) of $$\int_1^\infty\left(\int_0^1\frac{dx}{-1+x^3+y^2}\right)dy.\tag{1}$$

I would like to know if is it possible to get and justify an approximation using analysis (only is required two or three right decimal digits, and after the theoretical justification you can invoke calculations using your computer). I know that $$\int_1^\infty\frac{dy}{-1+x^3+y^2}=\frac{\arctan\left(\sqrt{x^3-1}\right)}{\sqrt{x^3-1}},$$ and I don't know if it is a good idea to use the Taylor series of the inverse tangent function here.

Second, if we calculate the closed-form of the inner integral in $(1)$ I think that also should be difficult to get approximation using Taylor series.

Question. Provide a reasoning to get the integer part and the first two decimal digits (after the decimal point) of $(1)$. Many thanks.

It is obvious that our integral is greater than $1$.

Answer 1

It's easier to expand the integrand of the double integral and integrate that term by term. We can write this as a geometric series:

$$f(x,y) = \sum_{n=0}^{\infty}\frac{\left(1-x^3\right)^n}{y^{2n+2}}$$

The integral over $x$ can be evaluated by substituting $u = 1-x^3$, the integral over $u$ then becomes $\frac{1}{3}\int_{0}^{1}u^n (1-u)^{-2/3}du$, this can be expressed in terms of a beta function: $\frac{1}{3} \dfrac{n!\left(-\frac{2}{3}\right)!}{(n+\frac{1}{3})!} = \left(\frac{1}{3}\right)! \dfrac{n!}{(n+\frac{1}{3})!}$

The integral over $y$ yields a factor of $\dfrac{1}{2n +1}$, therefore the integral can be written as :

$$\sum_{n=0}^{\infty}\left(\frac{1}{3}\right)! \frac{n!}{(2n+1)(n+\frac{1}{3})!}\tag{1}$$

The summation converges very slowly, but it's easy to deal with that by treating the tail of the summation using Stirling's formula for the factorials and then applying the Euler–Maclaurin formula to approximate the summation over the tail. Note that Stirling's formula can itself be derived using the Euler–Maclaurin formula.

Stirling's formula yields the following expansion for the summand in (1):

$$\left(\frac{1}{3}\right)! \frac{n^{-1/3}}{(2n+1)}\left(1-\frac{2}{9 n} +\frac{7}{81 n^2}-\frac{70}{2187 n^3}+\frac{182}{19683 n^4}-\frac{364}{177147 n^5}+\cdots\right)\tag{2}$$

We can now sum the first, say, $20$ terms of the summation in (1) and then approximate the summation from $n = 21$ to infinity by replacing the summand in (1) by the expansion (2), and then compute that summation of the tail, term by term by applying the Euler–Maclaurin formula to each term. This yields the result $2.30112685455\cdots$ for the summation, where the given $11$ decimals are correct. With more terms of the summation combined with more terms of the Euler–Maclaurin formula it's easy to get a large number of decimals.

Answer 2

I know functions which are much more pleasant to expand as Taylor series around $x=0$ than $$f=\frac{\arctan\left(\sqrt{x^3-1}\right)}{\sqrt{x^3-1}}$$ but a CAS gave $$f=\frac{1}{2} (2 \log (2)-3 \log (x))+\frac{1}{4} x^3 (-3 \log (x)-1+2 \log (2))+\frac{1}{32} x^6 (-18 \log (x)-7+12 \log (2))+\frac{1}{192} x^9 (-90 \log (x)-37+60 \log (2))+O\left(x^{12}\right)$$ which can write $$f=-\left(\frac{3}{2}+\frac{3 x^3}{4}+\frac{9 x^6}{16}+\frac{15 x^9}{32}+O\left(x^{12}\right) \right)\log(x)+$$ $$\left(1+\frac{x^3}{2}+\frac{3 x^6}{8}+\frac{5 x^9}{16}+O\left(x^{12}\right) \right)\log(2)-$$ $$\left(\frac{x^3}{4}+\frac{7 x^6}{32}+\frac{37 x^9}{192}+O\left(x^{12}\right) \right) $$ in which the coefficients reveal some interesting patterns; however, these coefficients vary very slowly and let then suppose a very slow convergence for the integral.

The only "problematic" integral is $$\int_0^1x^n\log(x)\,dx=-\frac{1}{(n+1)^2}\qquad \text{if} \qquad \Re(n)>-1$$ Using the above expansion and this last result, we find $$\int_0^1 \frac{\arctan\left(\sqrt{x^3-1}\right)}{\sqrt{x^3-1}}\,dx=\frac{68209+56910 \log (2)}{47040}\approx 2.28861$$ while the numerical integration of the double integral would give $\color{red}{2.30109}$.

Adding the next term to the expansion would lead to $$\frac{61016077+52186680 \log (2)}{42398720}\approx 2.29227$$ Adding the next term to the expansion would lead to $$\frac{7763990653+6763367520 \log (2)}{5427036160}\approx 2.29444$$ Adding the next term to the expansion would lead to $$\frac{8368328064511+7394510017440 \log (2)}{5877480161280}\approx 2.29585$$

Edit

Because $\lim_{x\to 0} \, f =\infty$ but $\lim_{x\to 1} \, f =1$, we can expect that the numerical integration will have some difficulties. To show it, we considered the two integrals $$g_1=\int_{10^{-k}}^1 \frac{\arctan\left(\sqrt{x^3-1}\right)}{\sqrt{x^3-1}}\,dx \qquad \text{and} \qquad g_2=\int_0^{1-{10^{-k}}} \frac{\arctan\left(\sqrt{x^3-1}\right)}{\sqrt{x^3-1}}\,dx $$ Below are reported the results $$\left( \begin{array}{ccc} k & g_1 & g_2 \\ 1 & 1.73637407698714 & 2.19584451172401 \\ 2 & 2.21011782123402 & 2.29107658640121 \\ 3 & 2.28857207445230 & 2.30012635428588 \\ 4 & 2.29952598877903 & 2.30102684955243 \\ 5 & 2.30093222130697 & 2.30111685450269 \\ 6 & 2.30110392707843 & 2.30112585455219 \\ 7 & 2.30112421944302 & 2.30112675455269 \\ 8 & 2.30112655626838 & 2.30112684455269 \\ 9 & 2.30112682127567 & 2.30112685355269 \end{array} \right)$$

Update

May be, it would have been simpler to make the expansion around $x=1$ to get for example $$f=1-y+\frac{4 y^2}{5}-\frac{62 y^3}{105}+\frac{3 y^4}{7}-\frac{123 y^5}{385}+\frac{1256 y^6}{5005}-\frac{1044 y^7}{5005}+\frac{15429 y^8}{85085}+O\left(y^{9}\right)$$ where $y=x-1$ and integrate from $-1$ to $0$.

For an expansion up to $O\left(y^{n+1}\right)$, the results would have been $$\left( \begin{array}{cc} n & \text{result} \\ 10 & 2.28861 \\ 20 & 2.29585 \\ 30 & 2.29848 \\ 40 & 2.29930 \\ 50 & 2.29976 \\ 60 & 2.30014 \\ 70 & 2.30032 \\ 80 & 2.30045 \\ 90 & 2.30058 \\ 100 & 2.30065 \end{array} \right)$$