Computing $f(x) + f\left(\frac{1}{x}\right)$ where $f(x)=\int_{1}^{x}\frac{ \log(t)}{1+t}\,\mathrm{d}t$

Question

If $$f(x)=\int_{1}^{x}\frac{ \log(t)}{1+t}\,\mathrm{d}t,$$ then what is the value of $f(x) + f\left(\dfrac{1}{x}\right)$?

I tried integration by parts, but it did not lead to a meaningful answer. Any kind of substitution also did not seem to work.

I was intending to solve the definite integral first and then proceed by substituting $x$ with $\dfrac{1}{x}$. But I was unable to solve the definite integral by any of the above mentioned methods.

Kindly suggest a correction to my approach, or if I am going right, a method to solve the definite integral would be appreciated.

Answer 1

Note that $$f(1/x)=\int_{1}^{1/x}\frac{\log(t)}{1+t}dt.$$ By substituting $y=1/t$, we get $$f(1/x) = \int_{1}^{x}\frac{-\log(y)}{1+1/y}\cdot \frac{-1}{y^2}dy = \int_{1}^{x}\frac{\log(t)}{t(1+t)}dt.$$ Therefore, $$f(x)+f(1/x) = \int_{1}^{x}\frac{\log(t)}{1+t}\left(1+\frac{1}{t}\right)dt=\int_{1}^{x}\frac{\log(t)}{t}dt=\cdots$$

Answer 2

Don't compute $f$ first. Get the second term by substituting $t=1/u$, then sum. You should find it simplifies nicely.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{f}\pars{x} & \equiv \int_{1}^{x}{\ln\pars{t} \over 1 + t}\,\dd t = -\int_{-1}^{-x}{\ln\pars{-t} \over 1 - t}\,\dd t = \ln\pars{1 + x}\ln\pars{x} - \int_{-1}^{-x}{\ln\pars{1 - t} \over t}\,\dd t \\[5mm] & = \ln\pars{1 + x}\ln\pars{x} + \int_{-1}^{-x}\mrm{Li}_{2}'\pars{t}\,\dd t = \ln\pars{1 + x}\ln\pars{x} + \mrm{Li}_{2}\pars{-x} - \mrm{Li}_{2}\pars{-1} \\[5mm] & = \bbx{\ln\pars{1 + x}\ln\pars{x} + \mrm{Li}_{2}\pars{-x} + {\pi^{2} \over 12}} \end{align}

Answer 4

With $u=1/t$, we have \begin{align*} F(x)&=\int_{1}^{1/x}\dfrac{1}{u}\dfrac{\log u}{1+u}du, \end{align*} so \begin{align*} F(x)+F(1/x)&=\int_{1}^{1/x}\left(\dfrac{1}{t}+1\right)\dfrac{\log t}{1+t}dt\\ &=\int_{1}^{1/t}\dfrac{\log t}{t}dt\\ &=\dfrac{1}{2}(\log t)^{2}\bigg|_{t=1}^{t=1/x}\\ &=\dfrac{1}{2}(\log x)^{2}. \end{align*}

Answer 5

Let $$ F(x)=f(x) + f\left(\dfrac{1}{x}\right).$$ Then $F(1)=0$ and $$ F'(x)=f'(x) + f'\left(\dfrac{1}{x}\right)(-x^{-2})=\frac{\ln(x)}{ 1 + x}-\frac{\ln(\frac1x)}{ 1 + \frac1x}\frac1{x^2}=\frac{\ln(x)}{ 1 + x}+\frac{\ln(x)}{ x(1 + x)}=\frac{\ln(x)}{x}. $$ So $$ F(x)=F(1)+\int_1^x\frac{\ln(t)}{t}dt=\frac12\ln^2x.$$