Integral of $x^{\sqrt x}$

Question

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.

Answer 1

$$x^{\sqrt{x}} = e^{\sqrt{x}\ln(x)} = \sum_{k = 0}^{+\infty} \frac{(\sqrt{x}\ln(x))^k}{k!}$$

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)