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How can I compute this integral: $I=\int\frac{1-(au+1)\exp(-au)}{u^{2}}du$ $$Can anyone help me in this case?

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    Thank you for your answers!2012-11-18

2 Answers 2

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Seems you have

$I=\int\frac{1-axe^{-ax}-e^{-ax}}{x^2}dx=\int\frac{dx}{x^2}-a\int\frac{e^{-ax}}{x}du-\int\frac{e^{-ax}}{x^2}dx$

making parts for the third integral:

$u=e^{-ax}\;,\;u'=-ae^{-ax}\;\;;\;\;v'=\frac{1}{x^2}\;,\;v=-\frac{1}{x}$

we get

$\int\frac{e^{-ax}}{x^2}dx=-\frac{e^{-ax}}{x}-a\int\frac{e^{-ax}}{x}dx$

so that we get:

$I=\frac{e^{-ax}-1}{x}+C$

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$I(a)=\int\frac{1-(au+1)\exp(-au)}{u^{2}}du$ then $\dfrac{dI}{da} = \int \dfrac{u(au+1) \exp(-au) - u\exp(-au)}{u^2} du = \int a \exp(-au) du = - \exp(-ax) + c$ Hence, $I(a) = \dfrac{\exp(-ax)}{x} + \underbrace{ca}_{\text{constant}} + d(x)$ Further $I(0) = 0 \implies d(x) = - \dfrac1x$. Hence, $I(a)=\int\frac{1-(au+1)\exp(-au)}{u^{2}}du = \dfrac{\exp(-ax)-1}x + \text{ constant}$

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    This is less elementary of what I did but I like it...+12012-11-18