How to prove


Here is how I came up with this relation:

In this solution @Kemono Chen elegantly proved

$$\int_0^a\frac{\ln(1+ax)}{1+x^2}dx=\int_0^1\frac{a\ln(1+a^2x)}{1+a^2x^2}dx=\frac12\arctan a\ln(1+a^2)\tag{2}$$

and while trying to prove the identity in (2) starting from RHS, I ended up with the relation in (1). So any straightforward method to prove (1)? Plus any good applications of (1)?

The transformation of the integral in (2) was done by @Jack D'Aurizio here.

I will post my proof in the answer section soon and I am tagging "harmonic number" as the proof involves it in case you are curious. Thanks

UPDATE: If we let $\frac{1-x}{x}\mapsto x$ in (1) then combine with (2) we have

$$\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^\infty\frac{\ln x}{a^2+(1+x)^2}=\frac1{2a}\arctan a\ln(1+a^2)\tag{3}$$

  • 1
    $\begingroup$ About the UPDATE. We can also prove that: $$\int_0^\infty\frac{\ln x}{a^2+(1+x)^2}=\frac1{2a}\arctan a\ln(1+a^2)$$ By letting $\frac{{1+a^2}}{x}\to x$ to get: $$\int_0^\infty \frac{\ln \left(\frac{1+a^2}{x}\right)}{a^2+(1+x)^2}dx$$ And the result follows by adding this with the initial integral. In general, it's useful to let $\frac{a}{x}=t$ when we encounter: $$\int_0^\infty \frac{\ln x}{x^2+bx+a}dx$$ $\endgroup$
    – Zacky
    Aug 31, 2019 at 18:18
  • $\begingroup$ Very nice trick. $\endgroup$ Aug 31, 2019 at 18:24

2 Answers 2


In other words we want to show that: $$\color{blue}{\int_0^1 \frac{\ln\left(\frac{1-x}{1+a^2 x}\right)}{1+a^2 x^2}dx}=\color{red}{\int_0^1 \frac{\ln x}{1+a^2 x^2}dx}$$ This can be seen via the substitution: $$\frac{1-x}{1+a^2 x}=t\Rightarrow x=\frac{1-t}{1+a^2 t}\Rightarrow dx=-\frac{1+a^2}{(1+a^2t)^2}dt$$ $$\Rightarrow \color{blue}{\int_0^1 \frac{\ln\left(\frac{1-x}{1+a^2 x}\right)}{1+a^2 x^2}dx}=\int_0^1 \frac{\ln t}{1+a^2 \frac{(1-t)^2}{(1+a^2t)^2}}\frac{1+a^2}{(1+a^2t)^2}dt\overset{t=x}=\color{red}{\int_0^1 \frac{\ln x}{1+a^2 x^2}dx}$$

  • $\begingroup$ Very nice @Zacky this is the type of solution I'm looking for.. neat and short (+1) $\endgroup$ Aug 31, 2019 at 17:13
  • $\begingroup$ Thank you! I am still trying to find some aplications for $(1)$. Maybe this: math.stackexchange.com/q/3240779/515527 can be proved differently using this result. $\endgroup$
    – Zacky
    Aug 31, 2019 at 17:16
  • $\begingroup$ that's a nice application. We would like to see some applications for tough problems. I hope we can find some. $\endgroup$ Aug 31, 2019 at 17:22

In the post body we have

$$\int_0^1\frac{a\ln(1+a^2x)}{1+a^2x^2}dx=\frac12\arctan a\ln(1+a^2)\tag{*}$$

and from this solution we have

\begin{align} f(a)&=\frac12\arctan a\ln(1+a^2)=-\sum_{n=0}^\infty \frac{(-1)^n H_{2n}}{2n+1}a^{2n+1}\\ &=-\sum_{n=0}^\infty \frac{(-1)^n H_{2n+1}}{2n+1}a^{2n+1}+\sum_{n=0}^\infty \frac{(-1)^n }{(2n+1)^2}a^{(2n+1)}\\ &=\int_0^1a\ln(1-x)\sum_{n=0}^\infty(-a^2x^2)^n-\int_0^1a\ln x\sum_{n=0}^\infty(-a^2x^2)^n\\ &=\int_0^1\frac{a\ln(1-x)}{1+a^2x^2}\ dx-\int_0^1\frac{a\ln x}{1+a^2x^2}\ dx\\ &=\int_0^1\frac{a\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}\ dx\tag{**} \end{align}

From (*) and (**) we have



$-\frac{H_{2n+1}}{2n+1}=\int_0^1x^{2n}\ln(1-x)\ dx$

$\frac1{(2n+1)^2}=-\int_0^1 x^{2n}\ln x\ dx$


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