# Integral $I=\int_0^1 \frac{\ln x\arctan(ax)}{1+b^2x^2}dx$

Greetings I am trying to evaluate $$I=\int_0^1 \frac{\ln x\arctan(ax)}{1+b^2x^2}dx$$ Where $a$ and $b$ are positive numbers. My try was to derivate the integral with respect to $a$ in order to get: $$I'(a)=\int_0^1 \frac{\ln x}{(1+a^2x^2)(1+b^2x^2)}dx=\frac{1}{a^2-b^2}\left(\int_0^1\frac{a^2\ln x}{1+a^2x^2}dx-\int_0^1\frac{b^2\ln x}{1+b^2x^2}dx\right)$$ expanding into geometric series using: $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n$$ $$I'(a)=\frac{1}{a^2-b^2}\left(a^2\sum_{n=0}^{\infty}(-1)^na^{2n}\int_0^1x^{2n}\ln xdx-b^2\sum_{n=0}^{\infty}(-1)^nb^{2n}\int_0^1x^{2n}\ln xdx\right)$$ Using the following relation:$$I(k)=\int_0^1x^{2k}dx=\frac{1}{2k+1}\rightarrow I'(k)=\int_0^1 x^{2k}\ln xdx=\frac{-2}{(2k+1)^2}$$ gives: $$I'(a)=\frac{1}{a^2-b^2}\left(-2a\sum_{n=0}^{\infty}(-1)^n\frac{a^{2n+1}}{(2n+1)^2}+2b\sum_{n=0}^{\infty}(-1)^n\frac{b^{2n+1}}{(2n+1)^2}\right)$$ $$I'(a)=\frac{2}{a^2-b^2}\left(b\sin b-a\sin a\right)$$ And finally $$I=I(a)=2\int \frac{b\sin b- a\sin a}{a^2-b^2}da$$ Now I am stuck, could you help me finish this? And do I have any mistakes?

Edit: With the mistake pointed in the comments it yields to: $$I'(a)=\frac{2}{a^2-b^2}\left(\sum_{n=1}^{\infty}(-1)^n\frac{a^{2n}}{(2n)^2}-\sum_{n=1}^{\infty}(-1)^n\frac{b^{2n}}{(2n)^2}\right)=\frac{1}{2}\frac{1}{a^2-b^2}(Li_2(-a^2)-Li_2(-b^2))$$

• You missed an $x$ after you differentiated wrt $a$. Jun 1, 2018 at 12:41
• How did you obtain $\cos a-\cos b$ in the last line?
– user
Jun 1, 2018 at 14:41
• Why don't you start with a simple particular case like $\displaystyle \int_0^1 \frac{\ln x\arctan(x/2)}{1+x^2}dx$? See what you get to have then a better image of what to expect. Jun 1, 2018 at 15:42
• Note how I wrote the word "simple" above. I mean to calculate the integral for a certain particular case (not a trivial one) as I mentioned and see how the closed-form looks like (it might not look like a jewel, but eventually all depends on everyone's taste). Jun 1, 2018 at 17:54

For one, you're missing an $x$ in the numerator when you differentiating $\arctan ax$.

Let $F(a)$ denote our integral and differentiating with respect to the parameter gives

\begin{align*}F'(a) & =\int\limits_0^1 dx\,\frac {x\log x}{(1+a^2x^2)(1+b^2x^2)}\\ & =\frac {a^2}{a^2-b^2}\int\limits_0^1dx\,\frac {x\log x}{1+a^2x^2}-\frac {b^2}{a^2-b^2}\int\limits_0^1dx\,\frac {x\log x}{1+b^2x^2}\end{align*}

Both integrals can be evaluated, like you've done with your work, using the infinite geometric sequence. The first integral, with $a$, is equal to

\begin{align*}\int\limits_0^1dx\,\frac {x\log x}{1+a^2x^2} & =-\sum\limits_{n\geq1}\frac {(-a^2)^n}{(2n+2)^2}\\ & =-\frac 14\left[\sum\limits_{n\geq1}\frac {(-a^2)^{n-1}}{n^2}-1\right]\\ & =-\frac 14\left[-1-\frac {\operatorname{Li}_2(-a^2)}{a^2}\right]\\ & =\frac {1}{4}+\frac 1{4a^2}\operatorname{Li}_2(-a^2)\end{align*}

Similarly, the other integral is the exact same; but with $a$ replaced with $b$. Putting everything together

$$F'(a)=\frac 1{a^2-b^2}\left[\frac {a^2}4+\frac 14\operatorname{Li}_2(-a^2)-\frac {b^2}4-\frac 14\operatorname{Li}_2(-b^2)\right]$$

Now integrate that monster to retrieve $F(a)$. I will update this when I finish figuring out how to efficiently integrate $F'(a)$.

EDIT: There's no way.

• @Zacky Yes.$\phantom{}$ Jun 1, 2018 at 18:42
• @Zacky The dilogarithm starts at $n=1$$\operatorname{Li}_2(x)=\sum\limits_{\color{red}{n\geq1}}\frac {x^n}{n^2}$$Here, when you let$n=1\$, the sum starts at two so we add/subtract one to get$$\sum\limits_{n\geq1}\frac {(-a^2)^n}{(n+1)^2}=\sum\limits_{n\geq1}\frac {(-a^2)^{n-1}}{n^2}-1$$ Jun 1, 2018 at 22:33