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

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.

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, $$f(x)+f(1/x) = \int_{1}^{x}\frac{\log(t)}{1+t}\left(1+\frac{1}{t}\right)dt=\int_{1}^{x}\frac{\log(t)}{t}dt=\cdots$$
• Then how can i calculate $f(x)+f\bigg(\frac{1}{x}\bigg)$ – DXT Mar 27 '18 at 7:36
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*}
Don't compute $f$ first. Get the second term by substituting $t=1/u$, then sum. You should find it simplifies nicely.