# How do you calculate $\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{1}{2t}\arctan\frac{t^2}{\sqrt{3t^4-4}}dt$?

How do you calculate $$\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{1}{2t}\arctan\frac{t^2}{\sqrt{3t^4-4}}dt?$$ Mathematica fails to do it.

• Start with $$t^2=u$$ and then $$\dfrac u{\sqrt{3u^2-4}}=\tan v\implies u^2=?$$ – lab bhattacharjee Jan 1 '17 at 8:05
• Maple also fails to do this. – Dr. Sonnhard Graubner Jan 1 '17 at 10:18
• I assume it can not be solved in such simple way...maybe we should use the technique of contour integral. – Jiyuan Zhang Jan 1 '17 at 11:09

Substitute $u=t^4$ to obtain $$\frac{1}{8} \int_{4/3}^2 \frac{\tan^{-1} \sqrt{\frac{u}{3u-4}}}{u} \ \mathrm{d}u$$ Consider $$I(a) = \int_{4/3}^2 \frac{\tan^{-1} \sqrt{\frac{a u}{3u-4}}}{u} \ \mathrm{d}u$$ which, when we take the derivative with respect to $a$, we obtain $$I'(a) = \int_{4/3}^2 \frac{1}{2 \sqrt{\frac{au}{3u-4}} (3u-4) \left(1+\frac{au}{3u-4}\right)}\ \mathrm{d}u$$

The integrand has antiderivative $$\frac{\sqrt{u} \left(\sqrt{a} \tan ^{-1}\left(\frac{\sqrt{a u}}{\sqrt{3 u-4}}\right)+\sqrt{3} \log \left(3 \sqrt{u}+\sqrt{9 u-12}\right)\right)}{(a+3) \sqrt{3 u-4} \sqrt{\frac{a u}{3 u-4}}}$$

Substitute $u=2$ to obtain $$\frac{\sqrt{a} \tan ^{-1}\left(\sqrt{a}\right)+\sqrt{3} \log \left(3 \sqrt{2}+\sqrt{6}\right)}{\sqrt{a} (a+3)}$$ Take the limit as $u \to \frac{4}{3}$ to obtain $$\frac{\frac{\sqrt{3} \log (12)}{\sqrt{a}}+\pi }{2 a+6}$$

The difference of these expressions is $$-\frac{\frac{\sqrt{3} \log \left(2-\sqrt{3}\right)}{\sqrt{a}}-2 \tan ^{-1}\left(\sqrt{a}\right)+\pi }{2 a+6}$$ which is the value of $I'(a)$.

Note also that $I(0) = 0$.

Now, we want $I(1)$; the resulting integral is something Mathematica can perform, and it outputs the following:

simp[a_] := -((Pi - 2 ArcTan[Sqrt[a]] + (Sqrt Log[2 - Sqrt])/Sqrt[a])/(6 + 2a))

DSolve[func'[a] == simp[a] && func == 0, func[a], a] /. a -> 1 // FullSimplify

Out= {{func ->
1/48 (48 Catalan +
I (3 \[Pi]^2 - 4 I \[Pi] Log[729/512 (7 + 4 Sqrt)] +
24 (PolyLog[2, -2 - Sqrt] - PolyLog[2, I (-2 + Sqrt)] +
PolyLog[2, -2 + Sqrt] -
PolyLog[2, -I (2 + Sqrt)])))}}


That is, the original integral has value $\frac{1}{8}$ times that, which is (on taking real parts, since I know the value is real) $$\frac{1}{96} \left(12 C+6 \Im\left(\text{Li}_2\left(i \left(-2+\sqrt{3}\right)\right)+\text{Li}_2\left(-i \left(2+\sqrt{3}\right)\right)\right)+\pi \log \left(\frac{729}{512} \left(7+4 \sqrt{3}\right)\right)\right)$$

1/96 (12 Catalan +
6 Im[PolyLog[2, I (-2 + Sqrt)] +
PolyLog[2, -I (2 + Sqrt)]] + \[Pi] Log[729/512 (7 + 4 Sqrt)])

• Thank you so much for your help! – Jiyuan Zhang Jul 3 '17 at 1:49

Let $I$ represent the original integral and $$I(a)=\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{1}{2t}\arctan\frac{at^2}{\sqrt{3t^4-4}}dt.$$ Clearly $I(0)=0$ and $I(1)=I$. Note \begin{eqnarray} I'(a)&=&\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{\partial}{\partial a}\frac{1}{2t}\arctan\frac{at^2}{\sqrt{3t^4-4}}dt\\ &=&\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{t \sqrt{3 t^4-4}}{\left(a^2+3\right) t^4-4}dt\\ &=&\frac14\int_{(4/3)^{1/2}}^{2^{1/2}}\frac{\sqrt{3 t^2-4}}{\left(a^2+3\right) t^2-4}dt\\ &=&\frac{a \arctan\left(\frac{a t}{\sqrt{3 t^2-4}}\right)+\sqrt{3} \log \left(\sqrt{9 t^2-12}+3 t\right)}{4 \left(a^2+3\right)}\bigg|_{(4/3)^{1/2}}^{2^{1/2}}\\ &=&-\frac{\pi a-2 a \arctan(a)+\sqrt{3} \left(\log (6)-2 \log \left(3+\sqrt{3}\right)\right)}{8 \left(a^2+3\right)}. \end{eqnarray} So \begin{eqnarray} I(a) &=&-\int_0^1\frac{\pi a-2 a \arctan(a)+\sqrt{3} \left(\log (6)-2 \log \left(3+\sqrt{3}\right)\right)}{8 \left(a^2+3\right)}da\\ &=&\frac{1}{48} \pi \log \left(\frac{27}{64} \left(2+\sqrt{3}\right)\right)+\frac14\int_0^1\frac{a\arctan(a)}{a^2+3}da. \end{eqnarray} Now we solve \begin{eqnarray} \int_0^1\frac{a\arctan(a)}{a^2+3}da. \end{eqnarray} Let $$J(b)=\int_0^1\frac{a\arctan(ab)}{a^2+3}da.$$ Then $$J'(b)=\int_0^1\frac{a^2}{(a^2+3)(1+a^2b^2)}da=\frac{6 \arctan (b)-\sqrt{3} \pi b}{6 \left(b-3 b^3\right)}.$$ So \begin{eqnarray} \int_0^1\frac{a\arctan(a)}{a^2+3}da&=&\int_0^1\frac{6 \arctan (b)-\sqrt{3} \pi b}{6 \left(b-3 b^3\right)}db\\ &=&\frac{1}{48} \left(8 C+\pi \log \left(64 \left(97-56 \sqrt{3}\right)\right)\right) \end{eqnarray} which is obtained from Mathematica. Here $C$ is Catalan constant. Thus $$I=\frac{1}{192} \left(8 C+\pi \left(\log \left(64 \left(97-56 \sqrt{3}\right)\right)+4 \log \left(\frac{27}{64} \left(2+\sqrt{3}\right)\right)\right)\right)$$

• Just to let you know, Mathematica thinks the integral $\int_0^1 \frac{a \tan^{-1}(a)}{a^2+3} \ \mathrm{d}a \simeq 0.0800804$, while the original integral is roughly $0.0497286$. – Patrick Stevens Jan 1 '17 at 14:38
• @PatrickStevens, I used Mathematica to do the calculation. – xpaul Jan 1 '17 at 21:59
• Thank you so much for your help! – Jiyuan Zhang Jul 3 '17 at 1:49

Let $I$ denote the integral. Then Mathematica gives

$$I = \frac{1}{24}G - \frac{3\pi}{32}\log 2 + \frac{\pi}{16}\log 3 \approx 0.0497285555762 \cdots,$$

where $G$ is the Catalan's constant. Currently I obtained the following representation

$$I = \frac{\pi}{16}\log(3/2) - \frac{1}{4} \int_{0}^{1} \frac{w \arctan w}{3-w^2} \, dw,$$

though I am not sure if this will lead me to a correct way. I will update my answer when I find a full solution.

• decomposing $\arctan$ into $\log$'s the integral should be computable as $\text{Li}_2+ \text{elementary garbage}...$ – tired Jan 2 '17 at 11:07
• @tired, Good suggestion, but I am afraid that it may lead to a computation similar to Patrick Stevens's one. I deliberately tried to avoid using polylogarithm, though it may be possible that $3$ in the denominator is so special that we eventually need to rely on them. – Sangchul Lee Jan 2 '17 at 11:12
• At least it will cause a lot of (boring) work which i personally don't think is worth the effort – tired Jan 2 '17 at 11:18
• @tired, I definitely agree. That is another reason I am reluctant to delve into that way. – Sangchul Lee Jan 2 '17 at 11:20