Double Exponentials  



I was thinking recently about composed functions, specifically of functions composed with themselves. Of course, I started with ex. This function composed with itself is eex; a double exponential. I then started thinking about ∫eex, and it sucked me in! I couldn't figure out a way to do this without putting it in extremely complicated terms, functions and identities. I've looked at many calculators to do this, and all of them have functions and identities to automatically solve it. Here are some of the strategies I used:

Using -i = i-1, I found that eex = cos(sin(-ix) - i cos(-ix)) + i sin(sin(-ix) - i cos(-ix)). I then made the substitution u = sin(-ix) - i cos(-ix), du = u dx. This lead me to the expression ∫cos(u)/u du + i∫sin(u)/u du. I couldn't figure out either, and it turns out that they have their own special function (the trigonometric integrals [found here]).

Using -i = i-1, I found that eex = cos(-iex) + i sin(-iex). I made the substitution u = -iex, du = u dx. Unsurprisingly, this lead me to the same thing as last time.

I tried making the substitution u = ex, du = u dx, getting me the integral ∫eu/u du, which lead me to yet another specialized function, the exponential integral.

As you can see, nothing really worked out. I also tried complexifying the integral [found here], integral calculators, wolfram alpha, and many more things, but it never worked out.

I then went on to look at other functions composed with themselves, and it seems that many of them are very hard to integrate over. My challenge for you is to integrate a function composed with its self that isn't a polynomial, easily simplified, or purposefully easy. Good luck.

That one guy that's always on


I've found something cool:

The ∫ln(ln(x)) dx can be simplified by the substitution u = ln(x) to ln(u)eu du, which can then be simplified by the substitution y = ln(u) to ∫yeyeey, which is a tower of exponentials.

That one guy that's always on


Please Login or Register