# Weird indefinite integral homework questions

I'm solving a couple of integration problems using the method of changing variables, and would like assistance with two particular problems that I can't seem to solve. I completed rest of the problems in this problem set without much effort, but these two seem impossible.

I've tried changing a few different variables in both problems, and I tried to calculate the solution with Wolfram Alpha, but neither of those had any avail.

$$\int x^{e^{x^2}}~dx$$

and

$$\int\frac{dx}{x+\ln^2x}$$

are the problems that I'm trying to solve. Any help is much appreciated.

• They are not solvable in terms of elementary functions. And this is not a hint, it's a fact. However, I'll have some fun in writing possible approaches in a while! – Von Neumann Feb 7 '16 at 12:42
• @KimPeek. I wonder if they could be expressed in terms of any function. – Claude Leibovici Feb 7 '16 at 12:46
• Are you really supposed to find the anti-derivatives? – Adrian Feb 7 '16 at 13:08
• @adjan Yes. They're both part of a 20 question problem set. "Integrate the following by the method of changing variables" is what the question says. – rob1194 Feb 7 '16 at 13:14
• @rob1194 Are those exercises taken from a specific book or from lecture notes? – Von Neumann Feb 7 '16 at 13:16

This should be a comment, but it would be too long for the comments section, so I'll write it here. Please do not vote up / down unless you find I wrote a real malarkey. In that case, I'll provide to delete this or to improve it.

As we said in the comments, there are no expressions in terms of simple and elementary functions, and also most probably there are no answers in terms of any functions, as @Claude Leibovici stated.

The other problem is that your integrals are undefined, so either numerical approaches are available. I'll provide to show you some substitutions to make things clearer.

1

In this integral you might naively try the substitution

$$e^{x^2} = t ~~~~~ x^2 = \ln(t) ~~~~~ x = \sqrt{\ln(t)} ~~~~~ \text{d}t = 2t\ \sqrt{\ln(t)}$$

gaining

$$\frac{1}{2}\int\ \frac{(\ln(t))^{t/2 - 1/2}}{t}\ \text{d}t$$

and since it involves the form $t^{-1}[F(t)]^t$ the possibilities it can be integrated are like zero. I don't think one may find a more suitable expression (despite the fact it makes me to think about the Logarithmic Integral Special Function, but it's more complex since it has a $t$ in the exponent itself).

2

For this one, it could be good to use (for $t > 0$)

$$x = e^t ~~~~~~~ \text{d}x = e^t\ \text{d}t$$

to get

$$\int\frac{e^t}{e^t + t^2}\ \text{d}t$$

We may collect the $e^t$ factor to have

$$\int\frac{\text{d}t}{1 + t^2e^{-t}}$$

Which could be treated in several ways, if the integral had extrema. Having not, we cannot say that much.

Notice also that this substitution works for $t > 0$ only, so other possible ways are opened.

• Huh. Well we haven't even mentioned nonelementary integrals in class, so I suppose I'll just skip the questions. Thank you for taking the time to write the answer. :) – rob1194 Feb 7 '16 at 13:18
• @rob1194 You're welcome! Please, I'm really curious about. Try to ask your professor about those, and let us know!! Were they take from a book? Or from lecture notes? – Von Neumann Feb 7 '16 at 13:24
• Not sure, we just got the worksheet from the school's website. I skimmed through the book and didn't find them. I'll ask about them tomorrow and report back. – rob1194 Feb 7 '16 at 13:34

For $\int x^{e^{x^2}}~dx$ ,

$\int x^{e^{x^2}}~dx$

$=\int e^{e^{x^2}\ln x}~dx$

$=\int\left(1+\sum\limits_{m=1}^\infty\dfrac{e^{mx^2}\ln^mx}{m!}\right)~dx$

Let $u=\ln x$ ,

Then $x=e^u$

$dx=e^u~du$

$\therefore\int\left(1+\sum\limits_{m=1}^\infty\dfrac{e^{mx^2}\ln^mx}{m!}\right)~dx$

$=x+\int\sum\limits_{m=1}^\infty\dfrac{u^me^ue^{me^{2u}}}{m!}~du$

$=x+\int\sum\limits_{m=1}^\infty\sum\limits_{n=0}^\infty\dfrac{m^nu^me^{(2n+1)u}}{m!n!}~du$

$=x+\sum\limits_{m=1}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=0}^m\dfrac{(-1)^{m-k}m^nu^ke^{(2n+1)u}}{(2n+1)^{m-k+1}n!k!}+C$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)

$=x+\sum\limits_{m=1}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=0}^m\dfrac{(-1)^{m-k}m^nx^{2n+1}\ln^kx}{(2n+1)^{m-k+1}n!k!}+C$

For $\int\dfrac{dx}{x+\ln^2x}$ ,

Let $u=\ln x$ ,

Then $x=e^u$

$dx=e^u~du$

$\therefore\int\dfrac{dx}{x+\ln^2x}=\int\dfrac{e^u}{e^u+u^2}~du$

Case $1$: $u^2e^{-u}\leq1$ , i.e. $\dfrac{\ln^2x}{x}\leq1$

Then $\int\dfrac{e^u}{e^u+u^2}~du$

$=\int\dfrac{1}{1+u^2e^{-u}}~du$

$=\left(1+\int\sum\limits_{n=1}^\infty(-1)^nu^{2n}e^{-nu}\right)~du$

$=u+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^{n+1}(2n)!u^ke^{-nu}}{n^{2n-k+1}k!}+C$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)

$=\ln x+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^{n+1}(2n)!\ln^kx}{n^{2n-k+1}k!x^n}+C$