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Is it possible to represent this integral in terms of elementary functions?

$$\int \textrm{ln}(1+e^{x^{2}})dx$$

I saw a challenge on this site (integral challenge) and still can not figure out how to solve it (if it's solvable).

Addition

The full integral (the one I want to calculate) is $$\int \left [\textrm{ln}(1+e^{x^{2}})+2\frac{[e^{x^{2}}(2x^{2}-1)-1]}{(e^{x^{2}}+1)^{2}}\right ]dx$$

Perhaps combining the two terms gives rise to an elementary primitive.

otreblig
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  • This post about elementary integrals may be helpful https://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral – JB1 Jun 30 '17 at 02:34
  • For $|x|\gg 0$ the integrand function essentially is $x^2$, so the primitive is close to $\frac{x^3}{3}$. Then you may use Leucippus' approach with a minor twist to write to full asymptotic expansion through the error function. – Jack D'Aurizio Jun 30 '17 at 04:28

1 Answers1

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One method may be to expand the logarithm as seen by: \begin{align} \int \ln(1 + e^{x^2}) \, dx &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \, \int e^{n \, x^2} \, dx \\ &= \frac{\sqrt{\pi}}{2} \, \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \, \sqrt{n}} \, erfi(\sqrt{n} \, x), \end{align} where $erfi(x)$ is the imaginary error function.

Leucippus
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  • That's interesting. The full integral have a second term, as you can see in the link. I was thinking of doing some combination between the ln and this term to get a primitive. – otreblig Jul 01 '17 at 15:39