# How to evaluate $\int_0^{\frac{\pi}{4}} \tan(x) \ln^2(\sin(4x)) \, dx$?

Question: How to evaluate $$\int_0^{\frac{\pi}{4}} \tan(x) \ln^2(\sin(4x)) \, dx?$$

### My attempt

We will denote the main integral as $$\Omega$$. $$\Omega=\int_0^{\frac{\pi}{4}} \tan(x) \ln^2(\sin(4x)) \, dx= \int_0^{\frac{\pi}{4}} \tan(x) \left[\left(\ln 2 + \ln(\sin(2x)) + \ln(\cos(2x))\right)^2\right] dx$$

We start by using the logarithmic property for the argument $$\sin(4x)$$ $$\ln^2(\sin(4x)) = \left(\ln 2 + \ln(\sin(2x)) + \ln(\cos(2x))\right)^2$$

Expanding the square, we obtain: $$\ln^2(\sin(4x)) = \ln^2(2) + \ln^2(\sin(2x)) + \ln^2(\cos(2x)) + 2 \ln 2 \ln(\sin(2x)) + 2 \ln 2 \ln(\cos(2x)) + 2 \ln(\sin(2x)) \ln(\cos(2x))$$

Substituting this back into the integral, we get: $$\Omega = \int_0^{\frac{\pi}{4}} \tan(x) \left[\ln^2(2) + \ln^2(\sin(2x)) + \ln^2(\cos(2x)) + 2 \ln 2 \ln(\sin(2x)) + 2 \ln 2 \ln(\cos(2x)) + 2 \ln(\sin(2x)) \ln(\cos(2x))\right] dx$$

Can someone help me proceed further, or suggest any better ideas? Thanks!!

• Mathematica says: $$\frac{7 \zeta (3)}{4}+\frac{7 \ln ^3(2)}{6}-\frac{7}{24} \pi ^2 \ln (2)$$ Commented May 24 at 8:54
• Maple computes: $\frac76 \ln^32+\frac74 \zeta (3)-\frac{7\pi^2}{24} \ln2$. Commented May 24 at 11:30
• In the same vein as Bertrand87's answer, substitute $x\to\frac\pi4-x$ and average, then $x\to\arctan x$:$$I=\frac12\int_0^\tfrac\pi4\frac{\sec^2x}{1+\tan x}\log^2(\sin(4x))\,dx=\frac12\int_0^1\log^2\left(\frac{4x\left(1-x^2\right)}{\left(1+x^2\right)^2}\right)\,\frac{dx}{1+x}$$ Commented May 24 at 22:29

(Not a complete answer but too long for a comment):

$$\begin{eqnarray*} \Omega = \int_0^{\frac{\pi}{4}} \tan(x) \ln^2(\sin(4x))dx &=& \frac{1}{2}\int_0^{\frac{\pi}{2}} \tan\left(\frac{w}{2}\right)\ln^2(\sin(2w))dw\\ &=& \frac{1}{2}\int_0^{\frac{\pi}{2}} \tan\left(\frac{w}{2}\right)\ln^2(2\sin w \cos w)dw\\ &=& \frac{1}{2} \int_{0}^{\frac{\pi}{2}}\frac{\sin w \ln^2(2\sin w \cos w)}{1+\cos w}dw\\ \end{eqnarray*}$$

Now do the substitution $$\cos w \mapsto t$$. You will get:

$$\begin{eqnarray*} \Omega &=& \frac{1}{2}\int_{0}^{1} \frac{\ln^2\left(2 t \sqrt{1 - t^2}\right)}{1 + t}dt\\ &=& \frac{\ln^2(2)}{2}\int_{0}^{1} \frac{1}{1 + t}dt + \ln(2)\int_{0}^{1} \frac{\ln(t)}{1 + t}dt + \frac{1}{2}\int_{0}^{1}\frac{\ln^2(t)}{1 + t}dt +\frac{\ln(2)}{2}\int_{0}^{1} \frac{\ln(1 - t)}{1 + t}dt + \frac{\ln(2)}{2}\int_{0}^{1} \frac{\ln(1 + t)}{1 + t}dt + \frac{1}{2}\int_{0}^{1}\frac{\ln(t) \ln(1 - t)}{1 + t}dt + \frac{1}{2}\int_{0}^{1}\frac{\ln(t) \ln(1 +t)}{1 + t}dt + \frac{1}{8}\int_{0}^{1}\frac{ \ln^2(1 - t)}{1 + t}dt+\frac{1}{8}\int_{0}^{1}\frac{ \ln^2(1 +t)}{1 + t}dt+\frac{1}{4}\int_{0}^{1}\frac{ \ln(1 +t)\ln(1-t)}{1 + t}dt\\ \end{eqnarray*}$$

Some of these integrals are trivial, and others can be evaluated by consulting classic books such as (Almost) Impossible Integrals, Sums, and Series or Inside Interesting Integrals, etc., or by looking for through this forum or AoPS. After a long and tedious calculation, which I do not even intend to put here, you will arrive at this result:

$$\Omega = -\frac{7}{24}\pi^2\ln(2) + \frac{7}{6}\ln^3(2) + \frac{7}{4}\zeta(3)$$

Just to give a more detailed outline of the solution to this integral, here is the closed-form solution of the integrals that I think are the hardest, if that helps a little:

$$\begin{eqnarray*} \int_{0}^{1}\frac{\ln(t)\ln(1+t)}{1+t}dt &=& -\frac{\zeta(3}{8}\\ \int_{0}^{1}\frac{\ln(t)\ln(1-t)}{1+t}dt &=& \frac{13}{8}\zeta(3) - \frac{\pi^2}{4}\ln(2)\\ \int_{0}^{1} \frac{\ln^2(1-t)}{1+t}dt &=& \frac{21}{12}\zeta(3) + \frac{1}{3}\ln^3(2)-\frac{1}{6}\pi^2\ln(2)\\ \int_{0}^{1} \frac{\ln(1-t)\ln(1+t)}{1+t}dt &=& \frac{1}{8}\zeta(3) + \frac{1}{3}\ln^3(2)-\frac{1}{12}\pi^2\ln(2) \end{eqnarray*}$$

• very nice answer!! thanks a lot Commented May 25 at 4:17

Changing $$u=\tan(2x)$$ in the proposed integral, it comes \begin{align} \Omega&=\int_0^{\frac{\pi}{4}} \tan(x) \ln^2(\sin(4x)) \, dx\\ &=\frac12\int_0^\infty\frac{u}{(1+u^2)(1+\sqrt{1+u^2})}\ln^2\left( \frac{2u}{1+u^2} \right)\,du \end{align} Now, with $$u=\sinh t$$, $$$$\Omega=\frac12\int_0^\infty\frac{\sinh t}{\cosh t\,(1+\cosh t)}\ln^2\left( \frac{2\sinh t}{\cosh^2t} \right)\,dt$$$$ Representing $$$$\ln^2(z)=\left.\frac{\partial^2}{\partial s^2}z^s\right|_{s=0}$$$$ the integral reads after swapping derivation and integration operations (which is easily justified) $$$$\Omega=\left.\frac{\partial^2}{\partial s^2}2^{s-1}\int_0^\infty\frac{\sinh^{s+1} t}{\cosh^{2s+1} t\,(1+\cosh t)}\right|_{s=0}\,dt$$$$ Now, expanding $$(1+\cosh t)^{-1}$$, swapping summation and integration gives \begin{align} \Omega&=\left.\frac{\partial^2}{\partial s^2}2^{s-1}\sum_{n=0}^{\infty}(-1)^n \int_0^\infty \frac{\sinh^{s+1} t}{\cosh^{2s+n+2}t}\,dt\right|_{s=0} \end{align} The integral can be expressed as a Beta function (see Gradshteyn and Ryzhik, 3.512.2): $$$$\int_0^{\infty} \frac{\sinh ^\mu x}{\cosh ^\nu x} d x=\frac{1}{2} \mathrm{~B}\left(\frac{\mu+1}{2}, \frac{\nu-\mu}{2}\right) \quad[\operatorname{Re} \mu>-1, \quad \operatorname{Re}(\mu-\nu)<0]$$$$ then \begin{align} \Omega&=\left.\frac{\partial^2}{\partial s^2}2^{s-2}\sum_{n=0}^{\infty}(-1)^n \operatorname B\left( s/2+1,s/2+n/2+1/2 \right) \right|_{s=0}\\ &=\left.\frac{\partial^2}{\partial s^2}2^{s-2}\Gamma(s/2+1)\sum_{n=0}^{\infty}(-1)^n \frac{\Gamma\left(s/2+n/2+1/2 \right)}{\Gamma\left(s+n/2+3/2 \right)} \right|_{s=0}\\ &=\left.\frac{\partial^2}{\partial s^2}2^{s-2}\Gamma(s/2+1) \left[\frac{\Gamma(s/2+1/2)}{\Gamma(s+3/2)}{}_2F_1\left( s/2+1/2,1;s+3/2;1 \right)- \frac{\Gamma(s/2+1)}{\Gamma(s+2)}{}_2F_1\left( s/2+1,1;s+3/2;1 \right) \right]\right._{s=0}\\ &=\left.\frac{\partial^2}{\partial s^2}\frac{\left(\sqrt{\pi}\, \Gamma \left(s+1/2 \right)-\Gamma\left(s/2+1/2\right)^{2} 2^{s}\right) \Gamma \left(s/2\right)}{4 \Gamma\left(s+1/2 \right) \Gamma \left(s/2+1/2\right)}\right|_{s=0}\\ &=\frac74\zeta(3)+\frac76\ln^3(2)-\frac{7}{24}\pi^2\ln(2) \end{align} as expected from @MariuszIwaniuk. A CAS made the last differentiation but could be obtained tediously.

• That's a great answer! Thanks a lot. I'm going to try the last differentiation now. Commented May 25 at 4:19

Different method for calculating $$\Omega=\int_{0}^{1} \frac{\ln^2\left(2 t \sqrt{1 - t^2}\right)}{1 + t}dt$$ obtained by @Bertrand87 above:

$$\Omega=\int_{0}^{1} \frac{\ln^2\left(2 t \sqrt{1 - t^2}\right)}{1 + t}dt$$

$$=\int_{0}^{1} \frac{\ln^2(2 t) }{1 + t}dt+\int_{0}^{1} \frac{\ln(t)\ln(1-x^2)}{1 + t}dt+\ln(2)\int_{0}^{1} \frac{\ln(1-x^2)}{1 + t}dt+\frac14\int_{0}^{1} \frac{\ln^2(1-x^2)}{1 + t}dt.$$

The first integral is $$\frac32\zeta(3)-\ln(2)\zeta(2)+\ln^3(2)$$. The second integral can be found by using the generalization given in Theorem 3 of this preprint:

$$\int_0^1\frac{\ln^{q-1}(t)\ln(1-t^2)}{1+t}dt=(-1)^qq!\,\eta(q+1)-(-1)^q(q-1)!(2-2^{1-q})\ln(2)\zeta(q)$$ $$-(-1)^q(q-1)!\sum_{k=1}^{q-2}\zeta(q-k)\eta(k+1),\quad q=2,3,4,\cdots.$$

$$\Longrightarrow \int_0^1\frac{\ln(t)\ln(1-t^2)}{1+t}dt=\frac32\zeta(3)-\frac32\ln(2)\zeta(2).$$ As for the third and fourth integrals, we already have the generalization:

$$\int_0^1 \frac{\ln^{q-1}(1-t^2)}{1+t}dt=\frac{\sqrt{\pi}}{2q}\lim_{y\to 1}\frac{\partial^{q}}{\partial y^{q}}\frac{\Gamma(y)}{\Gamma(y-\frac12)},\quad q=2,3,4,\cdots$$

$$\Longrightarrow \int_0^1 \frac{\ln(1-t^2)}{1+t}dt=\ln^2(2)-\frac12\zeta(2);$$

$$\int_0^1 \frac{\ln^2(1-t^2)}{1+t}dt=2\zeta(3)-2\ln(2)\zeta(2)+\frac43\ln^3(2).$$

Combine all integrals, we get

$$\Omega=\frac72\zeta(3)-\frac72\ln(2)\zeta(2)+\frac73\ln^3(2).$$

• (+1) for the nice answer. Commented Jun 1 at 16:07