Evaluate the integral $\int_0^{\infty} \frac{b\ln{(1+ax)}-a\ln{(1+bx)}}{x^2} \,dx$ 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?
 A: 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
$$I
=ab \int_0^{\infty}\frac1x\left(  \frac{\ln(1+ax)}{ax} -\frac{\ln(1+bx)}{bx}\right)  \,dx 
=ab \ln \frac ba
$$
A: 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:
$$I(a,b)=\int_0^\infty\frac{b\ln(1+ax)-a\ln(1+bx)}{x^2}dx=ab(\ln b-\ln a)$$
A: 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
$$\int \ln(X+c)dX=$$
$$(X+c)\ln(X+c)-X$$
A: 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}
A: 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
$$\int_0^\infty \left(b f(a)- a f(b) \right) dx=a b \log \left(\frac{b}{a}\right)$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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$\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}}}}$

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}

\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}
