As the title states, I'm having quite a lot of trouble with the integral:

$$\int_{-\ln 2}^{\ln 2} \frac{(\sin x+x)^{1/3}}{e^x}dx$$

The problem is that my standard (admittedly sparse) repertoire of tricks seems to have no effect at all. I can't think of any clever substitution that would work, and it looks like a nasty integral all-round.

I was hoping to apply some symmetry arguments (especially considering the limits of integration), but the exponential function is neither even nor odd, so I don't see how this would work.

I'm afraid I have absolutely no idea how to approach this.

  • $\begingroup$ Can I ask where this integral comes from? $\endgroup$ Jun 7, 2016 at 15:34
  • $\begingroup$ It's been on my todo list for months, so I can't remember exactly where from. I think it could be from an MIT integration bee, or perhaps a Putnam something or other. $\endgroup$ Jun 8, 2016 at 1:15

2 Answers 2


With that $(\sin(x)+x)^{1/3}$, it's very unlikely to have a closed form antiderivative, and I don't see any way to get help from a contour integration. Taking some advantage of symmetry, you can express your integral as $-2 \int_0^{\ln 2} (\sin(x)+x)^{1/3} \sinh(x)\; dx$. The numerical value is approximately $-.4758557412777$. The Inverse Symbolic Calculator doesn't find anything for this. So I would hazard to guess there is no closed form.


Use a substitution of $e^{-x}=u$

$\rightarrow -x=\ln{u}$

$\rightarrow -e^{-x}=du$

Exponential gets cancelled in integral

$-\int [\sin {(\ln {u})}+\ln {u}] du$

This gives (I am omitting unnecessary details)

$ u\cos{\ln{u}}+u.\ln{u}- \int u.(\frac{1}{u}) du$

$= u\cos{\ln{u}}+u.\ln{u}- u $

Replacing $u=e^{-x}$


Dear, I have omitted some obvious details, and I guess you should be able to carry forward. Should you still have problems, you can inbox.

Someone tells me Ihave missed the $\frac{1}{3}$.

So starting from

$-\int [\sin {(\ln {u})}+\ln {u}]^\frac{1}{3} du$

It can be rewritten as



If we consider $[(\frac{\sin{(\ln{u})}}{\ln{u}}+1)]^{\frac{1}{3}}$, by a Mac Laurin series, we find that this limit is not defined. It is divergent. Hence a product having this component will also be divergent.

This integral cannot be calculated manually.

  • 1
    $\begingroup$ I think you missed the 1/3. $\endgroup$ Jun 8, 2016 at 5:21
  • $\begingroup$ @ Euler...IS_ALIVE, thanks for informing. I made some changes, but can't get beyond this. I shall look at it, when I am a bit free. $\endgroup$
    – Tosh
    Jun 8, 2016 at 6:17

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