# If If it is convergent, evaluate it?

## ${\int}_{0}^{\infty} 17 s {e}^{- 10 s} \mathrm{ds}$

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Dec 3, 2017

$\frac{17}{10} ^ 2$

#### Explanation:

Considering

$I = {\int}_{0}^{\infty} c s {e}^{- \alpha s} \mathrm{ds}$ we have

$\frac{d}{\mathrm{ds}} \left(s {e}^{- \alpha s}\right) = {e}^{- \alpha s} - \alpha s {e}^{- \alpha s}$ or

$c \frac{d}{\mathrm{ds}} \left(s {e}^{- \alpha s}\right) = c {e}^{- \alpha s} - \alpha c s {e}^{- \alpha s}$ or

${\left(c s {e}^{- \alpha s}\right)}_{0}^{\infty} = {\int}_{0}^{\infty} c {e}^{- \alpha s} \mathrm{ds} - \alpha I$ and then

$I = \frac{1}{\alpha} \left({\int}_{0}^{\infty} c {e}^{- \alpha s} \mathrm{ds} - {\left(c s {e}^{- \alpha s}\right)}_{0}^{\infty}\right)$ but

${\int}_{0}^{\infty} c {e}^{- \alpha s} \mathrm{ds} = - \frac{1}{\alpha} c {\left({e}^{- \alpha s}\right)}_{0}^{\infty}$ and then

$I = \frac{1}{\alpha} \left(- \frac{1}{\alpha} {\left(c {e}^{- \alpha s}\right)}_{0}^{\infty} - {\left(c s {e}^{- \alpha s}\right)}_{0}^{\infty}\right)$ or

I = 1/alpha(-1/alphac(0-1)) -c(0-0)) = c/alpha^2

then

${\int}_{0}^{\infty} 17 {e}^{- 10 s} \mathrm{ds} = \frac{17}{10} ^ 2$

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