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I have encountered this integral and me and my colleagues cannot figure it out. I did prove that it is actually integrable (it is boundless for a certain $x_0$ as I will explain later), but I couldn't find a good way to solve it.

The integral is $$ \int^1_{-1}\:\frac{1}{\sqrt[3]{e^x+x}}\:\mathrm{d}x $$

What I found already:

  1. By Intermediate Zero Theorem on $(-1,0)$ there is a negative $x_0$ that nullifies the denominator. Thus, there the function is not defined and unbounded. We need to check that it is still integrable (improperly). Thus, we use Taylor Polynomial and we find that the denominator is asynptotic to $\sqrt[3]{(e^{x_0}+1)(x-x_0)}$. Since $\frac{1}{\sqrt[3]{x-x_0}}$ is integrable for $x\to x_0$, we can proceed to finding the value.
  2. Some euristic considerations on this part, which is the one I couldn't figure out. Those extrema bring to mind even/odd functions. This said, that, as well as substitution, seems complex since we cannot solve algebrically $e^x-x=t$. My hope was to divide the integral in "odd_function (morbid) + even_function (easy to integrate)" and nullify the odd one by using extrema.

Thank you.

Simone Licciardi
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    0.9153(appx) By Integral calculator. –  Jan 10 '23 at 22:41
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    Is there a reason you believe this has a closed form? Is it from some kind of competition? – Ninad Munshi Jan 10 '23 at 22:42
  • We can try bounds rather that finding the actual one and try to minimize the error. –  Jan 10 '23 at 22:42
  • No closed form exist Using digamma, gamma, etc –  Jan 10 '23 at 22:43
  • It was on a book. Unless its a mistake on the author side... – Simone Licciardi Jan 10 '23 at 22:44
  • https://www.integral-calculator.com/ –  Jan 10 '23 at 22:52
  • Try this it might help you –  Jan 10 '23 at 22:53
  • Even it has no particular Antiderivative. –  Jan 10 '23 at 22:53
  • Using some other Techinques Like Converting integral to Sum there might be the solution but I couldn't want to approach that. –  Jan 10 '23 at 22:57
  • Thank you for your help. – Simone Licciardi Jan 10 '23 at 22:59
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    @JeffreyHarkness: Wolfram Alpha calculates it as 0.915366 (which agrees with the value in the first comment). – NickD Jan 11 '23 at 00:16
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    @NickD, thanks for commenting. I meant to say: by solving $e^x+x=0$ numerically, I find that the vertical asymptote occurs near $x=−0.56714$. Using numerical approximation in R, I find that the positive integral value from the asymptote to 1 is somewhere near 1.661607. – Jeffrey Harkness Jan 11 '23 at 00:37
  • Can you tell us what book it came from? – polychroma Jan 11 '23 at 02:50
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    Maybe we can use the Taylor series $$\frac{1}{(1-x)^{1/3}}=\sum_{n=0}^\infty\frac{\Gamma(n+1/3)}{\Gamma(1/3)}\frac{x^n}{n!}$$ – polychroma Jan 11 '23 at 02:53
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    If you're willing to use Lambert-W functions (the inverse to the function $xe^x$), you can solve $e^x+x=u$ as $x=u-W(e^u)$. For instance, using this transforms the integral into $$\int_{1/e-1}^{e+1}\frac{u^{-1/3}}{1+W(e^u)}\text{ d}u$$ – Phobo Havuz Jan 11 '23 at 03:54
  • @PhoboHavuz. I am sorry ! I did not see your comment which is the last of a long list. I shall edit and give you the credit of this substitution. Cheers :-) – Claude Leibovici Jan 11 '23 at 06:08
  • Yea unfortunately even mathematica fails to spit out a closed form and I cannot see any way to contour integrate or series bash this into something analytical so I think the best we can do is just an approximation – Max0815 Jan 11 '23 at 13:15

1 Answers1

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$$I=\int^{+1}_{-1}\:\frac{dx}{\sqrt[3]{e^x+x}}$$

As @PhoboHavuz proposed in comments, $$e^x+x=t \implies x=t-W\left(e^t\right) \implies dx=\frac{dt}{1+W\left(e^t\right)}$$ $$I=\int_{\frac{1-e}{e}}^{1+e} \frac{dt}{\sqrt[3]{t} \,\left(1+W\left(e^t\right)\right)}$$

Now, for approximation, we could replace $\frac{1}{1+W\left(e^t\right)}$ by its $[n,n]$ Padé approximant $P_n$ built around $t=0$.

The simplest is $$P_1=\frac 1{(\Omega +1)}\,\frac {2 (\Omega +1)^2-t }{2 (\Omega +1)^2+ (2 \Omega -1)t}$$

To give an idea of their accuracy, the norms $$\Phi_n=\int_{\frac{1-e}{e}}^{1+e}\left(\frac{1}{1+W\left(e^t\right)}-P_n \right)^2\,dt$$

$$\left( \begin{array}{cc} n & \Phi_n \\ 1 & 1.164\times 10^{-2} \\ 2 & 7.318\times 10^{-5} \\ 3 & 8.019\times 10^{-7} \\ 4 & 9.955\times 10^{-9} \\ 5 & 1.284\times 10^{-10} \\ 6 & 1.638\times 10^{-12} \\ \end{array} \right)$$

Using these Padé approximants, after factorizations of numerator and denominator, the integrand write $$ \frac{1}{\sqrt[3]{t}}\,\frac {a_n \,\prod_{i=1}^n (t-r_i) } {b_n \,\prod_{i=1}^n (t-s_i) } $$ Using partial fractions, it becomes $$ \frac{a_n}{b_n}\,\frac{1}{\sqrt[3]{t}}\,\Bigg[1+\sum_{i=1}^n \frac{A_i}{t-s_i}\Bigg]$$ which is more than simple to integrate, leading, as usual, to a sum of logarithms and a sum of arctangents.

As a function of $n$, some decimal representations of analytical results $$\left( \begin{array}{cc} n & \text{result} \\ 1 & 0.808336 \\ 2 & 0.922701 \\ 3 & 0.914691 \\ 4 & 0.915434 \\ 5 & 0.915359 \\ 6 & 0.915367 \\ \end{array} \right)$$

Edit

Building the Padé approximants is quite simple starting from the expansion around $t=0$ $$\frac{1}{1+W\left(e^t\right)}=\sum_{n=0}^\infty (-1)^n \frac {P_n} {n!\,(\Omega+1)^{2n+1} }\,t^n$$

$$\left( \begin{array}{cc} n & P_n \\ 0 & 1 \\ 1 & \Omega \\ 2 & -\Omega+2 \Omega^2 \\ 3 & \Omega-8 \Omega^2+6 \Omega^3 \\ 4 & -\Omega+22 \Omega^2-58 \Omega^3+24 \Omega^4 \\ 5 & \Omega-52 \Omega^2+328 \Omega^3-444 \Omega^4+120 \Omega^5 \\ 6 & -\Omega+114 \Omega^2-1452 \Omega^3+4400 \Omega^4-3708 \Omega^5+720 \Omega^6 \\ \end{array} \right)$$

which show patterns which would be interesting to explore (the first one is trivial, the next are clearly related to Eulerian numbers).

Claude Leibovici
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