TITLE: Integral of $x^{\sqrt x}$
QUESTION [5 upvotes]: I'm trying to find this integral:
$$\int x^\sqrt x \, dx $$
Wolframalpha gave me an integral. (So it does exist)
I tried integration by parts & tried converting it to $$ e^{\sqrt x \ln(x)} $$ then expanding $e$ by its summation notation.

REPLY [5 votes]: $$x^{\sqrt{x}} = e^{\sqrt{x}\ln(x)} = \sum_{k = 0}^{+\infty} \frac{(\sqrt{x}\ln(x))^k}{k!}$$
Thence you get
$$\sum_{k = 0}^{+\infty} \frac{1}{k!} \int \left(\sqrt{x}\ln(x)\right)^k\ \text{d}x$$
A repeated integration by parts gives:
$$\int \left(\sqrt{x}\ln(x)\right)^k\ \text{d}x = \Gamma\left[1 + \frac{k}{2},\ -\left(1 + \frac{k}{2}\right)\ln(x)\right]\ln^{1 + k/2}(x)\left(-\left(1 + \frac{k}{2}\right)\ln(x)\right)^{-1 - k/2}$$
So in the end we have
$$\sum_{k = 0}^{+\infty} \frac{1}{k!}\left(\Gamma\left[1 + \frac{k}{2},\ -\left(1 + \frac{k}{2}\right)\ln(x)\right]\ln^{1 + k/2}(x)\left(-\left(1 + \frac{k}{2}\right)\ln(x)\right)^{-1 - k/2}\right)$$
Gamma Function
More here about the Gamma function
Incomplete Gamma Function
More here about the incomplete Gamma Function (which is the used one)