# Finding Integrals of Non-Elementary Functions

1. Sep 11, 2014

### Gwozdzilla

1. The problem statement, all variables and given/known data
We know that F(x) = $\int^{x}_{0}e^{e^{t}} dt$ is a continuous function by FTC1, though it is not an elementary function. The Functions $\int\frac{e^{x}}{x}dx$ and $\int\frac{1}{lnx}dx$ are not elementary funtions either but they can be expressed in terms of F.

a) $\int^{2}_{1}\frac{e^{x}}{x}dx$

b)$\int^{3}_{2}\frac{1}{lnx}dx$

2. Relevant equations
I only used and u-substitution in my attempt at a solution, but I suppose integration by parts and trigonometric substitution is fair game as well.

Integration by parts: $\int udv$ = uv - $\int vdu$

3. The attempt at a solution

a) $\int^{2}_{1}\frac{e^{x}}{x}dx$

Let x = et
dx =etdx

$\int\frac{e^{e^{t}}e^{t}}{e^{t}}dt$

$\int^{2}_{1}e^{e^{t}} dt$

Is this answer correct? If it isn't, where did I go wrong? If it is, how do I put it in terms of F(x)?

b)$\int^{3}_{2}\frac{1}{lnx}dx$

let u = ln(x)
du = 1/x dx
xdu = dx
eudu = dx

$\int\frac{e^{u}}{u}dx$

Which this is just part a again.. but am I allowed to do u substitution that way by plugging in u into my du line since it ended up being in terms of an x I didn't have to substitute with?

2. Sep 11, 2014

### LCKurtz

Your limits on the last integral are incorrect; those are the $x$ limits. You need the new $t$ limits that correspond to them.

Once you have any integral of the form $\int_a^b e^{e^t}~dt$ you should be able to write it as $F(b) - F(a)$. Do you see why?

3. Sep 11, 2014

### Gwozdzilla

So to find the correct variables...

2 = et
ln(2) = t
and
1 = et
ln(1) = t = 0

$\int^{ln(2)}_{0}e^{e^{t}} dt$

and the final answer would be F(ln2) - F(0) because FTC1 says

F(x) = $\int^{b}_{a}f(x) dx$ = F(b) - F(a)

Right?

4. Sep 11, 2014

### LCKurtz

and what is F(0)?

Yes and no. How can you have x on the left and no x on the right?

5. Sep 11, 2014

### Gwozdzilla

I'm not sure what F(0) is. I feel inclined to say that F(0) = e, but that's without evaluating $\int e^{e^{t}}$. But I don't know how to evaluate that integral. Do you want me to try to evaluate it or are you looking for a different answer? I'm confused.

For your second question... x isn't on the right side of that equation because a and b are being plugged in for x? I'm confused here too.

6. Sep 11, 2014

### LCKurtz

$F(x) = \int_0^x e^{e^t}~dt$. What do you get if you put $x=0$?

With that definition of $F(x)$ does it work out that $F(b)-F(a) = \int_a^be^{e^t}~dt$?

7. Sep 11, 2014

### Gwozdzilla

If x = 0, then we're taking the integral from 0 to 0, which would have to be 0. Right?

Does F(b) - F(a) = $\int^{b}_{a} e^{e^{t}} dx$? I don't see why not..

Is it supposed to be F(b)F'(b) - F(a)F'(a)? I know that happens every now and then, but I'm pretty sure that's only for the case in which I'm looking for the derivative of F(x).

8. Sep 12, 2014

### LCKurtz

Yes.

That's not much of an answer. Do the calculation. Use the formula for $F(x)$ to write down $F(b)$ and $F(a)$ and work it out.

9. Sep 12, 2014

### HallsofIvy

Staff Emeritus
Yes, that is correct. What LCKurtz was objecting to before, when you wrote
$$F(x)= \int_a^b e^{e^t} dt= F(b)- F(a)$$
was the "F(x)= " on the left. There is NO "x" in $\int_a^b e^{e^t} dt= F(b)- F(a)$ so it cannot be any function of x.

10. Sep 12, 2014

### nrqed

Watch out, on the right side of the last equation you should have $e^t dt$, not $e^t dx$