If f(x) = e^{-x} + 2e^{-2x} + 3e^{-3x} + dots | vedclass

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(B) The given series is $f(x) = \sum_{n=1}^{\infty} n(e^{-x})^n$. This is an arithmetico-geometric series of the form $\sum nx^n = \frac{x}{(1-x)^2}$

Vedclass · Accepted Answer

$\frac{1}{2}$

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(B) The given series is $f(x) = \sum_{n=1}^{\infty} n(e^{-x})^n$. This is an arithmetico-geometric series of the form $\sum nx^n = \frac{x}{(1-x)^2}$ where $x = e^{-x}$.Thus,$f(x) = \frac{e^{-x}}{(1-e^{-x})^2}$.We need to evaluate $I = \int_{\ln 2}^{\ln 3} \frac{e^{-x}}{(1-e^{-x})^2} \, dx$.Let $u = 1 - e^{-x}$. Then $du = e^{-x} \, dx$.When $x = \ln 2$,$u = 1 - e^{-\ln 2} = 1 - \frac{1}{2} = \frac{1}{2}$.When $x = \ln 3$,$u = 1 - e^{-\ln 3} = 1 - \frac{1}{3} = \frac{2}{3}$.So,$I = \int_{1/2}^{2/3} \frac{1}{u^2} \, du = \left[ -\frac{1}{u} \right]_{1/2}^{2/3}$.$I = -\left( \frac{1}{2/3} - \frac{1}{1/2} \right) = -\left( \frac{3}{2} - 2 \right) = -\left( -\frac{1}{2} \right) = \frac{1}{2}$.

If $f(x) = e^{-x} + 2e^{-2x} + 3e^{-3x} + \dots + \infty$,then $\int_{\ln 2}^{\ln 3} f(x) \, dx =$

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