Calculate the following, if it exists: $\int_{0}^{1} x^a(lnx)^mdx$ , where $a > -1$ and $m$ is a nonnegative integer.

We would like to evaluate 

$$ \int_{0}^{1} x^a (\ln x)^m \, dx $$

where $a > -1$ and $m$ is a nonnegative integer. This can be tackled by iintegration by parts. Let us consider $u = (\ln x)^m$ and $dv = x^a dx$. Then, $du = m (\ln x)^{m-1} \frac{1}{x} dx$ and $v = \frac{x^{a+1}}{a+1}$.

Integration by parts is given by the formula:

$$ \int u \, dv = uv - \int v \, du$$

Applying this formula, we have:

$$ \int_{0}^{1} x^a (\ln x)^m \, dx = \left. \frac{x^{a+1}}{a+1} (\ln x)^m \right|_0^1 - \int_{0}^{1} \frac{x^{a+1}}{a+1} m (\ln x)^{m-1} \frac{1}{x} \, dx$$

The boundary term at $x = 0$ vanishes because $x^{a+1} (\ln x)^m$ tends to $0$ as $x$ approaches $0$. At the upper limit $x = 1$, $(\ln x)^m$ is $0$ for all $m$, making the boundary term $\frac{x^{a+1}}{a+1} (\ln x)^m = 0$ there as well.

Thus, we are left with:

$$ \int_{0}^{1} x^a (\ln x)^m \, dx = -\frac{m}{a+1} \int_{0}^{1} x^a (\ln x)^{m-1} \, dx .      (*)$$

Using this formula recursively we get 
\[\int_{0}^{1} x^a (\ln x)^m \, dx = -\frac{m}{a+1} \int_{0}^{1} x^a (\ln x)^{m-1} \, dx\]
\[=(-\frac{m}{a+1})\cdot (-\frac{m-1}{a+1}) \int_{0}^{1} x^a (\ln x)^{m-1} \, dx\]
\[=(-\frac{m}{a+1})\cdot (-\frac{m-1}{a+1}) \dots (-\frac{1}{a+1})\int_{0}^{1} x^a (\ln x) \, dx\]
\[=(-1)^{m}\frac{m!}{(a+1)^m}\int_{0}^{1} x^a (\ln x)\, dx\]
\[=(-1)^{m}\frac{m!}{(a+1)^m} \big ( \left. \frac{x^{a+1}}{a+1} \ln x \right|_0^1 - \int_{0}^{1} \frac{x^{a+1}}{a+1}  \frac{1}{x} \, dx\big)\]
\[=(-1)^{m}\frac{m!}{(a+1)^m} \big (-\frac{1}{a+1}\int_0^{1} x^a dx \big)\]
\[=(-1)^{m}\frac{m!}{(a+1)^m} \big (-\frac{1}{(a+1)^2}dx\big)\]
\[=(-1)^{m}\frac{m!}{(a+1)^{m+2}}. \]