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On the entrance exam of the Tokyo Institute of Technology, the problem below was asked:

Find the integer part of a real number $\displaystyle \int^{2023}_{0} \dfrac{2}{x+e^x}dx$.

This problem asked that the entrance exam candidates use inequalities $\dfrac{2}{2e^x-1}\leq\dfrac{2}{x+e^x}\leq \dfrac{2}{e^x}$ or something similar to evaluate this value, as the indefinite integral cannot be written in elementary functions. Then a question occurred to me: Can this indefinite integral be written with some famous functions that are not elementary? Though I have no clue about this, I'm curious about this. I would appreciate it if you have any ideas. Thank you!

Arturo Magidin
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Jiner Soling
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    The integral is less than $$\int\limits_0^{2023}2e^{-x},dx =2(1-e^{-2023})$$ Therefore the integer part is equal $0$ or $1.$ – Ryszard Szwarc Sep 28 '23 at 17:37
  • The integral can only be calculated numerically , but to answer the question, the calculation of the integral is not necessary. – Peter Sep 28 '23 at 17:38
  • I would proceed by factoring $e^x$ out of the denominator, then rewrite the fraction using geometric series, and integrate term by term. You would likely need to use parametric differentiation to simplify the integral as I believe it would require several integration by parts. You may be able to write that resulting expression using special functions. – Cameron Williams Sep 28 '23 at 17:52
  • Wolfram Alpha knows many special functions and will give them as the results of integrals, but it can't do this one. That doesn't prove that it can't be done; it doesn't even prove that it can't be done using only special functions that Wolfram Alpha knows about. But it's a bad sign. – Toby Bartels Sep 28 '23 at 18:56
  • Don’t think this one has an antiderivative defined with typical special functions, since mathematica didn’t output anything – Captain Chicky Sep 28 '23 at 20:06
  • Wolfram Alpha uses the Risch-algorithm. If it finds no antiderivate , it can usually safely be assumed that there is none. But as said , we do not need the antiderivate anyway. – Peter Sep 29 '23 at 01:22

3 Answers3

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You can't integrate with elementary functions, but you can get a very accurate estimate.

$\int \frac {2}{x+e^x} \ dx\\ \int 2e^{-x} \frac {1}{1+xe^{-x}} \ dx\\ \int 2e^{-x} \sum_\limits{n=0}^\infty (-xe^{-x})^n \ dx\\ 2\int \sum_\limits{n=0}^\infty (-1)^nx^ne^{-(n+1)x} \ dx$

We can interchange the integral and summation when we have dominated convergence.

$2\sum_\limits{n=0}^\infty (-1)^n \int x^ne^{-(n+1)x} \ dx\\ \int x^ne^{-(n+1)x} \ dx = \frac {n!}{(n+1)^{n+1}}e^{-(n+1)x}P(x)$

Where $P(x)$ is a polynomial.

Evaluating at $x = 2023$ any polynomial will be trivial compared to the exponential factor that preceeds it.

$\int_0^{2023} x^ne^{-(n+1)x} \ dx$ very nearly equals $\int_0^{\infty} x^ne^{-(n+1)x} \ dx = \frac {n!}{(n+1)^{n+1}}$

In a small number of terms $(n+1)^{n+1} \gg n!$

$2(1 - \frac 1{4} + \frac {2}{27} - \frac {6}{256} + \cdots)\approx 1.613$
Which matches what we find from numerical integration.

https://www.wolframalpha.com/input?i=%5Cint%5B0to+2023%5D+++2%2F%28x+%2B+e%5Ex%29+dx

J.G.
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user317176
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There are already a few answers, but here is a simple series expansion substituting and using the Lambert W series:

$$e^x+x=u\implies x=f(u)du,f(u)=u+\sum_{n=1}^\infty \frac{(-1)^ne^{nu}n^{n-1}}{n!}$$

Now integrate:

$$\int\frac{dx}{e^x+x}=\int\frac{f’(u)}udx=\ln(u)+\sum_{n=1}^\infty\frac{(-1)^n n^n}{n!}\int\frac{e^{nu}}udu$$

Therefore the Ei function appears:

$$\int\frac1{e^x+x}dx=C+\ln(e^x+x)+\sum_{n=1}^\infty\frac{(-n)^n}{n!}\operatorname{Ei}(n(e^x+x))$$

shown here. This result works for about $x<-1$

Тyma Gaidash
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No, there is no function. But it should be smaller than $\int_0^{2023}\frac{2}{e^x}~dx$ and since this is less than $1$ the answer is $0$.

trula
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