# Finding $k$ such that $\int_0^{\pi/12} \ln(\tan (3x))\,\mathrm{d}x = k \int_0^{\pi/12} \ln(\tan(x))\,\mathrm{d}x$

$$\int_0^{\pi/12} \ln(\tan (3x))\,\mathrm{d}x = k \int_0^{\pi/12} \ln(\tan(x))\,\mathrm{d}x$$ What is $$k$$?

So I have tried almost every property so far and it has not been helpful, tried the property $$f(a+b-x)=f(x)$$

$$a$$ and $$b$$ being lower and upper limits of an integral for $$f(x)$$, tried rewriting $$\tan(3x)$$ to get $$\int_0^{\frac{\pi}{12}} \ln \left( \frac{3\tan(x)-\tan(x)^3} {1-3\tan(x)^2} \right)\mathrm{d}x$$ but then again I was stuck, how do we evalute this?

• Remark: Just like with $\ln$ common math functions render nicely with the extra backslash, i.e. $\tan$ gives $\tan$. Commented May 9, 2023 at 16:20
• Does this answer your question? Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$
– user1026811
Commented May 9, 2023 at 19:16
• Not that I know how to do this another way, but maybe you could use $\int_0^z\ln(\tan x)\,dx=z\,\ln(\tan z)-\operatorname{Ti}_2(\tan z)$. Commented May 10, 2023 at 0:55

If you use the complex representation of the arctangent function $$\tan^{-1}(t)=\frac{i}{2} \log{\left ( \frac{i+t}{i-t}\right )}$$ the antiderivative is not too difficult.

But, for the fun, consider the series expansions. In other words, $$I_1=\int_0^{\frac \pi {12}}\log (\tan (x))\,dx$$ $$\large\color{blue}{I_1=\frac{\pi}{12} \log \left(\frac{\pi }{12 e}\right)+\sum_{n=1}^\infty \frac{ \left(4^n-2\right) \pi^{2 n+1} \left| B_{2 n}\right| }{3^{2 n+1} 2^{2n+3} n \Gamma (2 n+2)}}$$ $$I_3=\int_0^{\frac \pi {12}}\log (\tan (3x))\,dx$$ $$\large\color{blue}{I_3=\frac{\pi}{12} \log \left(\frac{\pi }{4 e}\right)+\sum_{n=1}^\infty \frac{ \left(4^n-2\right) \pi ^{2 n+1} \left| B_{2 n}\right| }{3\, 2^{2 n+3}n \Gamma (2 n+2)}}$$

Limit the summation to $$p$$ terms of your choice and compute $$k_p=\frac {I_3^{(p)}}{I_1^{(p)}}$$ Notice that $$k_0=\frac{\log \left(\frac{\pi }{4 e}\right)}{\log \left(\frac{\pi }{12 e}\right)}=0.530543 \qquad \qquad \huge\color{red}{!!!}$$

Now a few more accurate values

$$\left( \begin{array}{cc} p & k_p \\ 0 & 0.530543 \\ 1 & 0.502892 \\ 2 & 0.500370 \\ 3 & 0.500056 \\ 4 & 0.500009 \\ 5 & 0.500002 \\ 6 & 0.500000 \\ \end{array} \right)$$

• thank you so much, will definitely learn this concept Commented May 16, 2023 at 17:25

Now you have found a method to solve the problem, I would like to give you the only method (other than seeking help of WoframAlpha) I know how to do this type of integral.

First express the function to be integrated as a series, e.g., $$\ln\left|\tan\left(x\right)\right|=\ln\left|x\right|+\dfrac{1}{3}x^2+\dfrac{7}{90}x^4+\dfrac{62}{2835}x^6+\dots+\dfrac{2^{2n}\left(2^{2n-1}-1\right)}{n\left(2n\right)!}B_nx^{2n} \quad\text{for}\quad 0\lt\left|x\right|\lt \dfrac{\pi}{2},$$ where $$B_n$$ denotes the Bernoulli numbers.

If you do this for both $$\space\ln\left|\tan\left(x\right)\right|\space$$ and $$\space\ln\left|\tan\left(3x\right)\right|,\space$$ and then evaluate the two definite integrals, you will get, $$k=0.5$$

This value of $$k$$ strongly suggests that there is a some kind of a trick to solve this problem. I believe that is what you have already found, so congratulations! I have no further ideas.

• thanks for this new concept and indeed k is 0.5 Commented May 9, 2023 at 17:52
• how do you compute this integral after doing the series expansion?
– user1026811
Commented May 9, 2023 at 18:04
• @GGplay You can integrate the series term by term up to certain number of terms.
– YNK
Commented May 9, 2023 at 18:24
• But how to get a closed form after integration term by term?
– user1026811
Commented May 9, 2023 at 18:52
• This is false, this integral can be solved analytically. Wolfram is not god.
– user1026811
Commented May 9, 2023 at 19:15