Evaluate the integral $\int_0^{\infty} \frac{b\ln{(1+ax)}-a\ln{(1+bx)}}{x^2} \,dx$

Question

Evaluate the following integral

$$I=\int_0^{\infty} \frac{b\ln(1+ax)-a\ln(1+bx)}{x^2} \,dx$$ with $\ a,b\in\mathbb{R},\ 0<a<b$.

My first attempt was to write $b\ln(1+ax)-a\ln(1+bx)$ as another integral, so I could substitute in the initial integral and then, by reversing the order of integration, try to calculate $I$.

I've tried writing $$b\ln(1+ax)-a\ln(1+bx)=\frac{ab}{t}\cdot\ln(1+tx)\Biggr|_{t=b}^{t=a}=ab\int_b^a\left(\frac{x}{t(1+tx)}-\frac{\ln(1+tx)}{t^2}\right)\,dt$$ but it doesn't seem to help me evaluating $I$.

What else could I try?

Answer 1

Apply the Frullani integral formula below

$$\int_0^\infty \frac{f(ax)-f(bx )}x =(f(0)-f(\infty))\ln\frac ba $$

with $f(x)=\frac {\ln(1+x)}x$ to obtain

$$I =ab \int_0^{\infty}\frac1x\left( \frac{\ln(1+ax)}{ax} -\frac{\ln(1+bx)}{bx}\right) \,dx =ab \ln \frac ba $$

Answer 2

Let's use differentiation under the integral to find the integral.

Let $I(a,b)=\int_0^\infty\frac{b\ln(1+ax)-a\ln(1+bx)}{x^2}dx$. Then, let's take the derivative with respect to $b$: $$\frac{\partial I}{\partial b}=\int_0^\infty\frac{\partial}{\partial b}\frac{b\ln(1+ax)-a\ln(1+bx)}{x^2}dx=\int_0^\infty \frac{\ln(1+ax)-\frac{ax}{1+bx}}{x^2}dx$$ Now, take the derivative with respect to $a$: \begin{align*}\frac{\partial^2 I}{\partial a \partial b}&=\int_0^\infty\frac{\partial}{\partial a} \frac{\ln(1+ax)-\frac{ax}{1+bx}}{x^2}dx\\&=\int_0^\infty\frac{\frac x{1+ax}-\frac x{1+bx}}{x^2}\\&=\int_0^\infty\frac 1x\cdot\frac{1+bx-1-ax}{(1+ax)(1+bx)}dx\\&=\int_0^\infty\frac{b-a}{(1+ax)(1+bx)}dx\\&=\int_0^\infty\frac{b}{1+bx}-\frac a{1+ax}dx\text{ by partial fractions}\\&=\ln{(1+bx)}-\ln{(1+ax)}\Big|_0^\infty\\&=\ln b - \ln a\end{align*} Where the last equality is left as an exercise for the reader :)

So, let's move backwards now, integrating with respect to $a$: $$\frac{\partial I}{\partial b}=a\ln b-a(\ln a -1)+c_1$$ for some real constant $c_1$. Then we integrate again with respect to $b$: $$I(a,b)=ab(\ln b-\ln a)+c_1b+c_2$$ for real constants $c_1,c_2$. Our goal now is to find those constants: Note that for non-negative real $a$, we have $I(a,a)=\int_0^\infty\frac{a\ln(ax+1)-a\ln(ax+1)}{x^2}dx=\int_0^\infty\frac 0{x^2}dx=0$. So, consider $I(1,1)$ and $I(2,2)$:

\begin{align*} I(1,1)&=0&=(1)(1)(\ln 1-\ln1)+c_1(1)+c_2&=c_1+c_2\\ I(2,2)&=0&=(2)(2)(\ln 2-\ln2)+c_1(2)+c_2&=2c_1+c_2 \end{align*} So $c_1=c_2=0$ and our final result is: $$I(a,b)=\int_0^\infty\frac{b\ln(1+ax)-a\ln(1+bx)}{x^2}dx=ab(\ln b-\ln a)$$

Answer 3

hint

With the substitution, $$t=\frac 1x$$

$$I=\int_0^{+\infty}(b\ln(t+a)-a\ln(t+b)+(a-b)\ln(t))dt$$

and $$\int \ln(X+c)dX=$$ $$(X+c)\ln(X+c)-X$$

Answer 4

Integrate by parts

\begin{align} & \int_0^{\infty} \frac{b\ln{(1+ax)}-a\ln{(1+bx)}}{x^2} dx\\\overset{IBP} = & ab\int_0^{\infty} \frac1x \left(\frac1{1+ax}- \frac1{1+b x} \right) dx = ab \int_0^{\infty} \left(-\frac a{1+ax}+ \frac b{1+b x} \right) dx \\ =& ab\ln\frac{1+bx}{1+ax}\bigg|_0^\infty = ab\ln\frac ba \end{align}

Answer 5

Consider first the antiderivative $$f(c)=\int \frac{\log (c x+1)}{x^2}\,dx$$ A first integration by parts gives $$f(c)=-\frac{\log (c x+1)}{x}+\int\frac{c}{x (c x+1)}\,dx$$ Partial fraction decomposition give $$\int\frac{c}{x (c x+1)}\,dx=\int \left(\frac{c}{x}-\frac{c^2}{c x+1} \right)\,dx=c \log (x)-c\log (c x+1)$$ As a total $$f(c)=c \log (x)-c \log (c x+1)-\frac{\log (c x+1)}{x}$$ Now, you consider $$b f(a)- a f(b)=\frac{a (b x+1) \log (b x+1)-b (a x+1) \log (a x+1)}{x}$$ Taking the limits at the bounds $$\int_0^\infty \left(b f(a)- a f(b) \right) dx=a b \log \left(\frac{b}{a}\right)$$

Answer 6

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\on{I}\pars{a,b} \equiv \left.\int_{0}^{\infty}{b\ln\pars{1 + ax} - a\ln\pars{1 + bx} \over x^{2}}\,\dd x \,\right\vert_{a,\, b\ \in\ \mathbb{R}_{\, >\, 0}}}}$

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Lets consider $\ds{\bbox[5px,#ffd]{\left.\int_{0}^{\infty} {\ln\pars{1 + px}\,x^{\nu - 2}}\,\,\dd x \,\right\vert_{% \substack{p\ >\ 0 \\[1mm] 0\ <\ \nu\ <\ 1}}}}$ which I'll evaluate by means of the Ramanujan's Master Theorem. Note that \begin{align} \ln\pars{1 + px} & = -\sum_{k = 1}^{\infty}{\pars{-px}^{k} \over k} \\[2mm] & = \sum_{k = 0}^{\infty}\braces{\color{red} {-\bracks{k \not= 0}\Gamma\pars{k}p^{k}}} {\pars{-x}^{k} \over k!} \end{align} Then, \begin{align} &\bbox[5px,#ffd]{\left.\int_{0}^{\infty} {\ln\pars{1 + px}\,x^{\pars{\color{red}{\nu - 1}} - 1}}\,\,\dd x \,\right\vert_{\substack{p\ >\ 0 \\[1mm] 0\ <\ \nu\ <\ 1}}} \\[5mm] = &\ \Gamma\pars{\nu - 1} \braces{-\bracks{1 - \nu \not= 0}\Gamma\pars{1 - \nu}p^{1 - \nu}} \\[5mm] = &\ -{\Gamma\pars{\nu} \over \nu - 1}\,\, {\Gamma\pars{1 - \nu}p^{1 - \nu}} = {p^{1 - \nu} \over 1 - \nu}\,{\pi \over \sin\pars{\pi\nu}} \end{align}

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\begin{align} \on{I}\pars{a,b} & \equiv \bbox[5px,#ffd]{\left.\int_{0}^{\infty}{b\ln\pars{1 + ax} - a\ln\pars{1 + bx} \over x^{2}}\,\dd x \,\right\vert_{a,\, b\ \in\ \mathbb{R}_{\, >\, 0}}} \\[5mm] & = \lim_{\nu\ \to\ 0^{+}}\,\,\bracks{% b\,{a^{1 - \nu} \over 1 - \nu}\,{\pi \over \sin\pars{\pi\nu}} - a\,{b^{1 - \nu} \over 1 - \nu} \,{\pi \over \sin\pars{\pi\nu}}} \\[5mm] & = \pi\lim_{\nu\ \to\ 0^{+}}\,\, {b\,a^{1 - \nu} - a\,b^{1 - \nu} \over \sin\pars{\pi\nu}} \\[5mm] = &\ \pi\lim_{\nu\ \to\ 0^{+}}\,\, {-b\,a^{1 - \nu}\,\ln\pars{a} + a\,b^{1 - \nu}\,\ln\pars{b} \over \cos\pars{\pi\nu}\pi} \\[5mm] = &\ \bbx{ab\ln\pars{b \over a}} \\ & \end{align}