Calculate the integral: integral from 1 to e x^m(ln x)^n

Calculate the integral: $\int_1^e x^m(\ln x)^n \, dx$, where $m$ and $n$ are non-negative integers.

We will use integration by parts repeatedly.

Step 1: Apply integration by parts

Let $u = (\ln x)^n$, so $u' = \frac{n(\ln x)^{n-1}}{x}$; and $dv = x^m \, dx$, so $v = \frac{x^{m+1}}{m+1}$ (assuming $m \neq -1$).

\[\int_1^e x^m(\ln x)^n \, dx = \left[\frac{x^{m+1}(\ln x)^n}{m+1}\right]_1^e - \frac{n}{m+1}\int_1^e x^m(\ln x)^{n-1} \, dx\]

Step 2: Evaluate the boundary term

\[\left[\frac{x^{m+1}(\ln x)^n}{m+1}\right]_1^e = \frac{e^{m+1} \cdot 1^n}{m+1} - \frac{1 \cdot 0^n}{m+1} = \frac{e^{m+1}}{m+1}\]

Step 3: Establish a recurrence relation

Let $I_{m,n} = \int_1^e x^m(\ln x)^n \, dx$. Then:

\[I_{m,n} = \frac{e^{m+1}}{m+1} - \frac{n}{m+1} I_{m,n-1}\]

Step 4: Find the base case $I_{m,0}$

\[I_{m,0} = \int_1^e x^m \, dx = \frac{e^{m+1} - 1}{m+1}\]

Step 5: Apply the recurrence relation

For $n=1$:

\[I_{m,1} = \frac{e^{m+1}}{m+1} - \frac{1}{m+1} \cdot \frac{e^{m+1} - 1}{m+1} = \frac{e^{m+1}\cdot m + 1}{(m+1)^2}\]

For $n=2$:

\[I_{m,2} = \frac{e^{m+1}}{m+1} - \frac{2}{m+1} I_{m,1} = \frac{e^{m+1}(1-m) - 2}{(m+1)^3}\]

Step 6: Recurrence formula summary

Applying the recurrence $n$ times links $I_{m,n}$ back to $I_{m,0}$ through the chain $I_{m,n}\to I_{m,n-1}\to\cdots\to I_{m,0}$; the recurrence relation itself, together with its base case, fully determines every value:

\[\boxed{I_{m,n} = \frac{e^{m+1}}{m+1} - \frac{n}{m+1} I_{m,n-1}, \quad I_{m,0} = \frac{e^{m+1} - 1}{m+1}}\]

Special cases:

For $m=0, n=1$:

\[I_{0,1} = \int_1^e \ln x \, dx = e - 1\cdot\frac{e-1}{1} = e - (e-1) = 1\]

For $m=1, n=1$:

\[I_{1,1} = \int_1^e x \ln x \, dx = \frac{e^2}{2} - \frac{1}{2}\cdot\frac{e^2-1}{2} = \frac{e^2+1}{4}\]

Final answer:

\[\boxed{\int_1^e x^m(\ln x)^n \, dx = \frac{e^{m+1}}{m+1} - \frac{n}{m+1}\int_1^e x^m(\ln x)^{n-1} \, dx}\]

with base case:

\[\boxed{\int_1^e x^m \, dx = \frac{e^{m+1} - 1}{m+1}}\]