1
$\begingroup$

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.

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

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[3] Log[2 - Sqrt[3]])/Sqrt[a])/(6 + 2a))

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

Out[80]= {{func[1] -> 
 1/48 (48 Catalan + 
  I (3 \[Pi]^2 - 4 I \[Pi] Log[729/512 (7 + 4 Sqrt[3])] + 
     24 (PolyLog[2, -2 - Sqrt[3]] - PolyLog[2, I (-2 + Sqrt[3])] +
         PolyLog[2, -2 + Sqrt[3]] - 
        PolyLog[2, -I (2 + Sqrt[3])])))}}

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[3])] + 
  PolyLog[2, -I (2 + Sqrt[3])]] + \[Pi] Log[729/512 (7 + 4 Sqrt[3])])
| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you so much for your help! $\endgroup$ – Jiyuan Zhang Jul 3 '17 at 1:49
2
$\begingroup$

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 $$ I=\frac{1}{192} \left(8 C+\pi \left(\log \left(64 \left(97-56 \sqrt{3}\right)\right)+4 \log \left(\frac{27}{64} \left(2+\sqrt{3}\right)\right)\right)\right)$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $0.0497286$. $\endgroup$ – Patrick Stevens Jan 1 '17 at 14:38
  • $\begingroup$ @PatrickStevens, I used Mathematica to do the calculation. $\endgroup$ – xpaul Jan 1 '17 at 21:59
  • $\begingroup$ Thank you so much for your help! $\endgroup$ – Jiyuan Zhang Jul 3 '17 at 1:49
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ decomposing $\arctan$ into $\log$'s the integral should be computable as $\text{Li}_2+ \text{elementary garbage}...$ $\endgroup$ – tired Jan 2 '17 at 11:07
  • $\begingroup$ @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. $\endgroup$ – Sangchul Lee Jan 2 '17 at 11:12
  • $\begingroup$ At least it will cause a lot of (boring) work which i personally don't think is worth the effort $\endgroup$ – tired Jan 2 '17 at 11:18
  • 1
    $\begingroup$ @tired, I definitely agree. That is another reason I am reluctant to delve into that way. $\endgroup$ – Sangchul Lee Jan 2 '17 at 11:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.