I need to evaluate the following integral.

$$\int_{0}^{2015}e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}dx. $$

But I still have no idea how to do it. Can anyone please give me some help? Thanks a lot.

  • 1
    $\begingroup$ Do you mean $e^{(e^{2015x})}$ or $(e^e)^{2015x}$? $\endgroup$
    – Pim
    Commented Feb 19, 2015 at 6:19
  • $\begingroup$ To get an idea calculate the derivatives of $e^{e^x}$ and $e^{e^{e^x}}$. $\endgroup$ Commented Feb 19, 2015 at 6:23
  • $\begingroup$ Is this a question from an on-going contest? $\endgroup$
    – JRN
    Commented Feb 19, 2015 at 8:51
  • $\begingroup$ It's $2015$ alright... $\endgroup$
    – AvZ
    Commented Feb 19, 2015 at 9:03

3 Answers 3


If you don't immediately see it, you can start with the substitution



We went from a product of four things to a product of three. And we can keep going with $u_2=e^{u_1}$ and $u_3=e^{u_2}$ to bring it down to $\frac1{2015}\int e^{u_3}du_3$. The hard part is the bounds. If you want to work things back, though, you can see that $u_3=e^{e^{e^{2015x}}}$. So we have


  • $\begingroup$ (+1) I just reviewed my answer after getting an upvote and realized I had misapplied a substitution. After fixing my answer, it is the same as yours. $\endgroup$
    – robjohn
    Commented Aug 19, 2022 at 13:03

Using the substitution $u_1=e^{2015x}$, then repeatedly using $u_{k+1}=e^{u_k}$, gives $$ \begin{align} \int_0^{2015}e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}\,\mathrm{d}x &=\frac1{2015}\int_1^{e^{2015^2}}e^{e^{e^{u_1}}}e^{e^{u_1}}e^{u_1}\,\mathrm{d}{u_1}\\ &=\frac1{2015}\int_e^{e^{e^{2015^2}}}e^{e^{u_2}}e^{u_2}\,\mathrm{d}{u_2}\\ &=\frac1{2015}\int_{e^e}^{e^{e^{e^{2015^2}}}}e^{u_3}\,\mathrm{d}{u_3}\\ &=\frac1{2015}\int_{e^{e^e}}^{e^{e^{e^{e^{2015^2}}}}}\,\mathrm{d}{u_4}\\ &=\frac1{2015}\left({e^{e^{e^{e^{2015^2}}}}}-e^{e^{e}}\right) \end{align} $$


Hint: Substitute $ u=e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x} $.

Then $$ du=2015e^{2015x+e^{e^{2015x}}+e^{2015x}}dx . $$

Therefore $$\int e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}dx=\int \frac{ue^{u-e^{e^{2015x}}}}{2015}dx \dots $$

  • 3
    $\begingroup$ Are you sure that you differentiated $u$ correctly? It seems to me that the product/chain rules would give more terms. $\endgroup$ Commented Feb 19, 2015 at 6:26
  • 1
    $\begingroup$ Right idea, wrong substitution. $\endgroup$
    – Mike
    Commented Feb 19, 2015 at 6:31
  • $\begingroup$ I edited my solution $\endgroup$
    – ASB
    Commented Feb 19, 2015 at 6:53

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