If $f(x) = e^{-x}+2e^{-2x}+3e^{-3x}+\cdots$ then find $\int_{\ln2}^{\ln3}f(x)dx$

Question

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.

Answer 1

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.

Answer 2

Integration term by term is valid because the convergence is uniform by the Weierstrass M-test, hence $$\int_{\ln2}^{\ln3}f(x)dx=\int_{\ln2}^{\ln3}\sum_{n=1}^\infty n e^{-nx}dx=\sum_{n=1}^\infty n\int_{\ln2}^{\ln3} e^{-nx}dx=\sum_{n=1}^\infty2^{-n}-3^{-n}=\frac{1}{2-1}-\frac{1}{3-1}=\frac{1}{2}.$$

Answer 3

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.