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I want to find the first 6 terms for the series expansion of this integral:

$$\int x^x~dx$$

My idea was to let: $$x^x=e^{x\ln x}$$

From that we have: $$\int e^{x\ln x}~dx$$

The series expansion of $e^x$ is: $$\sum\limits_{n=0}^\infty\frac{x^n}{n!}$$

Then we have:

$$\int e^{x\ln x}~dx=\int\sum\limits_{n=0}^\infty\frac{(x\ln x)^n}{n!}dx$$

I integrated the first 3 terms (since it was the easiest the work with): $$x+\frac{1}{4}x^2(2\ln x-1)+\frac{1}{54}x^3(9(\ln x)^2-6\ln x+2)+\cdots$$

Is there any other efficient way of getting the first 6 terms of the series expansion of the integral?

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You are in the right approach. Then you have to face $\int x^n(\ln x)^n~dx$ for non-negative integers $n$ :

$\int x^n(\ln x)^n~dx$

$=\int(\ln x)^n~d\left(\dfrac{x^{n+1}}{n+1}\right)$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{x^{n+1}}{n+1}d((\ln x)^n)$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)x}dx$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^n(\ln x)^{n-1}}{n+1}dx$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{n(\ln x)^{n-1}}{n+1}d\left(\dfrac{x^{n+1}}{n+1}\right)$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{x^{n+1}}{n+1}d\left(\dfrac{n(\ln x)^{n-1}}{n+1}\right)$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{n(n-1)x^{n+1}(\ln x)^{n-2}}{(n+1)^2x}dx$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{n(n-1)x^n(\ln x)^{n-2}}{(n+1)^2}dx$

$=\cdots\cdots$

$\vdots$

$\vdots$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\cdots\cdots+\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^n}-\int\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times1)x^n}{(n+1)^n}dx$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\cdots\cdots+\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^n}+\dfrac{(-1)^n(n(n-1)\cdots\cdots\times1)x^{n+1}}{(n+1)^{n+1}}+C$

$=\dfrac{(n+1)x^{n+1}(\ln x)^n}{(n+1)^2}-\dfrac{(n+1)nx^{n+1}(\ln x)^{n-1}}{(n+1)^3}+\cdots\cdots+\dfrac{(-1)^{n-1}((n+1)n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^{n+1}}+\dfrac{(-1)^n((n+1)n(n-1)\cdots\cdots\times1)x^{n+1}}{(n+1)^{n+2}}+C$

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

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

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

Hence $\int x^x~dx=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$

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Try taking a Taylor series for $\log x$, multiplying it by $x$, and plugging the result into the Taylor series for $e^x$. You can easily compute the first few terms.

Your choice of series for $\log x$ will affect where the result converges, so nothing will be valid everywhere. If this were a definite integral, it might be better.

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  • $\begingroup$ I kinda like the idea. But if I did that, wouldn't that depend on where the point $x_0$ is? $\endgroup$ – Mark Mar 18 '14 at 2:48
  • $\begingroup$ Yes it would. I havent tried it to see what happens. But the constant term will be arbitrary anyway since you are speifiying only an indefinite integral with no additional conditions. $\endgroup$ – MPW Mar 18 '14 at 11:28
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http://en.wikipedia.org/wiki/Exponential_formula

\begin{align} f(x) & = a_1 x + \frac{a_2}{2} x^2 + \frac{a_3}{6} x^3 + \frac{a_4}{24} x^4 + \frac{a_5}{120} x^5 + \frac{a_6}{720} x^6 + \cdots \\[10pt] e^{f(x)} & = 1 + a_1 x + \frac{a_2+a_1^2}{2} x^2 + \frac{a_3+3a_2a_1+ a_1^3}{6} x^3 + \frac{a_4 + 4a_3 a_1+ 3a_2^2 + 6a_2a_1^2 + a_1^4}{24} x^4 \\[8pt] & \phantom{{}={}} + \frac{a_5 + 5a_4 a_1 + 10a_3 a_2 + 10a_3 a_1^2 + 15 a_2^2 a_1 + 10 a_2 a_1^3 + a_1^5}{120} x^5 \\[8pt] & \phantom{{}={}} + \frac{a_6 + 6a_5 a_1 + 15 a_4 a_2 + 15 a_4 a_1^2 + 10a_3^2 + 60 a_3 a_2 a_1 + 20 a_3 a_1^3 + 15 a_2^3 + 45 a_2 a_1^4+ a_1^6}{720} x^6 \\[8pt] & \phantom{{}={}} + \cdots\cdots \end{align}

The patterns in the indices correspond to integer partitions and the coefficients count set partitions. For example, here is how we get the coefficient of $x^6$: $$ \begin{array}{l|c} \text{integer partition} & \text{number of set partitions} \\ \hline 6 & 1 \\ 5+1 & 6 \\ 4+2 & 15 \\ 4+1+1 & 15 \\ 3+3 & 10 \\ 3+2+1 & 60 \\ 3+1+1+1 & 20 \\ 2+2+2 & 15 \\ 2+2+1+1 & 45 \\ 2+1+1+1+1 & 15 \\ 1+1+1+1+1+1 & 1 \end{array} $$

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