# How to evaluate $\int_{0}^{2015}e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}dx$? [closed]

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.

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

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

$$u_1=e^{2015x},du_1=2015e^{2015x}$$

$$\int_0^{2015}e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}dx=\frac1{2015}\int_1^{e^{2015^2}}e^{e^{e^{u_1}}}e^{e^{u_1}}e^{u_1}du_1$$

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

$$\frac1{2015}\int_{e^{e^{e^0}}}^{e^{e^{e^{2015(2015)}}}}e^{u_3}du_3=\frac1{2015}(e^{e^{e^{e^{2015^2}}}}-e^{e^e})$$

• (+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.
– 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 $$\int e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}dx=\int \frac{ue^{u-e^{e^{2015x}}}}{2015}dx \dots$$

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