If $f(x) = e^{-x}+2e^{-2x}+3e^{-3x}+\cdots$ then find $\int_{\ln2}^{\ln3}f(x)dx$ Solve the following 
If $f(x) = e^{-x}+2e^{-2x}+3e^{-3x}+\cdots $
Then find $\int_{\ln2}^{\ln3}f(x)dx$
I don't have any idea.
 A: Integration term by term is valid because the convergence is uniform by the Weierstrass M-test, hence
$$\int_{\ln2}^{\ln3}f(x)dx=\int_{\ln2}^{\ln3}\sum_{n=1}^\infty n e^{-nx}dx=\sum_{n=1}^\infty n\int_{\ln2}^{\ln3} e^{-nx}dx=\sum_{n=1}^\infty2^{-n}-3^{-n}=\frac{1}{2-1}-\frac{1}{3-1}=\frac{1}{2}.$$
A: Let $x \in [\ln 2, \ln 3]$. Using the closed form of the geometric series, we get:
$$
g(x) = \sum_{n=1}^\infty e^{-nx} = \frac{1}{e^x - 1}
$$
Differentiate term by term to get:
$$
f(x) = -g'(x) = \sum_{n=1}^\infty n e^{-nx} = \frac{e^x}{(e^x - 1)^2}
$$
This is valid because convergence is uniform in both cases by the Weierstrass M-test.
Integrating the last function should be straightforward.
A: First off, your sum "to $\infty$" is suggestive, but bad notation. Don't write like that.
Start with (bear with me for a moment):
$$
\begin{align*}
\sum_{n \ge 0} z^n
  &= \frac{1}{1 - z}  \quad \lvert z \rvert < 1 \\
\sum_{n \ge 0} n z^{n - 1}
  &= \frac{d}{d z} \frac{1}{1 - z} = \frac{1}{(1 - z)^2} \\
\sum_{n \ge 1} n z^n
  &= \frac{z}{(1 - z)^2}
\end{align*}
$$
In your case, $z = e^{-x}$, so for the limits of your integral $e^{- \ln 2} = \frac{1}{2}$ and $e^{- \ln 3} = \frac{1}{3}$, which are comfortably inside the convergence range. So you can write:
$$
f(x) = \frac{e^{-x}}{(1 - e^{-x})^2}
$$
So your integral is:
$$
\begin{align*}
\int_{\ln 2}^{\ln 3} f(x) \, d x
  &= \int_{\ln 2}^{\ln 3} \frac{e^{-x} \, d x}{(1 - e^{-x})^2} \\
  &= - \int_2^3 \frac{du}{(1 - u)^2} \\
  &= \left. \frac{1}{1 - u} \right|_{u = 2}^3 \\
  &= \frac{1}{-2} - \frac{1}{-1} \\
  &= \frac{1}{2}
\end{align*}
$$
The manipulations on infinite series can be justified rigorously for power series inside their convergence radii.
