Indefinite integral of $x^x$

Question

I've seen many many questions on the internet with answer that it cannot be done with elementary functions. Now I did this integration myself and got a pretty nice result. Since I've seen so many answers telling it can't be done I have no idea where I might have gone wrong.

$$K=\int x^x \,dx$$ so $$\ln(K)=\ln\int x^x \,dx =\int \ln x^x \,dx = \int x\ln x \,dx\,.$$

Integration of $x\ln x$ can be done relatively simply.

$$\begin{align*} \int x\ln x \,dx &= (\ln x)\frac{x^2}{2}-\int \frac{x^2}{2}\frac{1}{x} \,dx\\ &= (\ln x)\frac{x^2}{2}-\frac{1}{2}\int x \,dx \\ &= \ln(x)\frac{x^2}{2}-\frac{1}{2}\frac{x^2}{2} \\ &= \ln(x)\frac{x^2}{2}-\frac{x^2}{4} \\ &= \frac{(\ln x^2)x^2-x^2}{4} \\ &= \ln K\end{align*}$$

Thus making $K$ to be equal to $K = e^{\frac{ln(x^2)*x^2-x^2}{4}+c}$

This seems rather elementary function to me, but I may have done some nasty mistake. Where did I go wrong?

Answer 1

Going from the second to the third line- that's where your mistake lies.

$$ \ln \left[\int x^x dx \right] \neq \int \ln(x^x) dx.$$

Remember that $\ln$ is a function and not a constant, so we cannot just pull it in and out the integral sign as and when we please; the only time we can do this is if we've got a constant.

Answer 2

That integral can be done by series. Write:

$$\int x^x \text{d}x = \int e^{x\ln(x)}\text{d}x = \sum_{k = 0}^{+\infty}\frac{1}{k!}\int \left(x\ln(x)\right)^k \text{d}x$$

and then repeat an integration by parts using

$$f'(x) = x^k ~~~~~~~ g(x) = \ln^k(x)$$

so

$$f(x) = \frac{x^{k+1}}{k+1} ~~~~~~~ g'(x) = \frac{k}{x}\ \ln^{k-1}(x)$$

After $n$ integrations by part you will obtain:

$$\int x^x \text{d}x = \sum_{k = 0}^{+\infty}\frac{1}{k!} \sum_{n = 0}^k (-1)^n \frac{x^{k+1}\cdot k! \cdot \ln^{k-n}(x)}{(k+1)^{n+1}\cdot (k-n)!}$$

or re arraging:

$$\int x^x \text{d}x = \sum_{k = 0}^{+\infty}\sum_{n = 0}^k (-1)^n \frac{x^{k+1} \cdot \ln^{k-n}(x)}{(k+1)^{n+1}\cdot (k-n)!}$$

Answer 3

Just have a look at : http://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function

$\int x^x \,dx = $ Sphd$(1 , x)$+ constant