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How do I prove that

$$\int_{0}^{\infty} du \left(\frac{u^{2}}{(u+a)^{3}} - \frac{u^{2}}{(u+b)^{3}}\right) = \ln \left(\frac{b}{a}\right)?$$

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    There is a general [Frullani's theorem](http://mathworld.wolfram.com/FrullanisIntegral.html), however in this case you can just simply find the antiderivative and take the limits. It's not that hard – 2017-02-10
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    What is the antiderivative in this case? – 2017-02-10
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    if you want to be fancy, integrate $(z^2(z+b)^3-z^2(z+b)^3)\frac{\log(z)}{(z+b)^3(z+b)^3}$ over a keyhole contour in the complex plane. Otherwise follow @Jack D'Aurizo's advice – 2017-02-10
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    $$\frac{u^2}{(u+a)^3}=\frac{u-a}{(u+a)^2}+\frac{a^2}{(u+a)^3}=\frac{1}{u+a}-\frac{2a}{(u+a)^2}+\frac{a^2}{(u+a)^3}$$ – 2017-02-10

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You can use the substitution $t = u + c$ $$ \int_{0}^{+\infty}\frac{u^2}{(u+c)^3}\,du = \int_{c}^{+\infty}\frac{t^2-2tc+c^2}{t^3}\,dt = \int_{c}^{+\infty}\frac{1}{t}\,dt - \int_{c}^{+\infty}\frac{2c}{t^2}\,dt + \int_{c}^{+\infty}\frac{c^2}{t^3}\,dt = \int_{c}^{+\infty}\frac{1}{t}\,dt + 2 - \frac{1}{2}$$

and plug in $a$ and $b$ to obtain $$\int_{0}^{+\infty} du \left(\frac{u^{2}}{(u+a)^{3}} - \frac{u^{2}}{(u+b)^{3}}\right) = \int_{a}^{+\infty}\frac{1}{t}\,dt-\int_{b}^{+\infty}\frac{1}{t}\,dt = \ln \left(\frac{b}{a}\right)$$

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    You need to be very careful with this last part: $$\int_{a}^{+\infty}\frac{1}{t}\,dt-\int_{b}^{+\infty}\frac{1}{t}\,dt = \ln \left(\frac{b}{a}\right)$$ – 2017-02-10
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    Yes, of course. It boils down to $$\int_{a}^{+\infty}\frac{1}{t}\,dt-\int_{b}^{+\infty}\frac{1}{t}\,dt = \int_{0}^{+\infty}(\frac{1}{u+a}-\frac{1}{u+b})\,du$$ – 2017-02-10
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    @AnubisBlack: technically speaking, no, it doesn't. In the LHS there is the difference between two diverging integrals, in the RHS there is a converging integral. It is like stating that $\infty-\infty = 1$. – 2017-02-10
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By integration by parts it boils down to $$ \int_{0}^{+\infty}\left(\frac{1}{u+a}-\frac{1}{u+b}\right)\,du = \log(b)-\log(a).$$

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\infty}\dd u \bracks{{u^{2} \over \pars{u + a}^{3}} - {u^{2} \over \pars{u + b}^{3}}} \\[5mm] = &\ \ \int_{0}^{\infty} \bracks{{u^{2} \over \pars{u + a}^{3}} - {1 \over u + a}}\dd u - \int_{0}^{\infty} \bracks{{u^{2} \over \pars{u + b}^{3}} - {1 \over u + b}}\dd u + \int_{0}^{\infty}\pars{{1 \over u + a} - {1 \over u + b}}\dd u = \\[5mm] = &\ \ \braces{\int_{0}^{\infty} \bracks{{u^{2} \over \pars{u + 1}^{3}} - {1 \over u + 1}}\dd u - \int_{0}^{\infty} \bracks{{u^{2} \over \pars{u + 1}^{3}} - {1 \over u + 1}}\dd u} + \left.\ln\pars{u + a \over u + b}\right\vert_{\ u\ =\ 0}^{\ u\ \to\ \infty} \\[5mm] = &\ \bbx{\ds{\ln\pars{b \over a}}} \end{align}