On the integrals $\int_0^{\frac{\pi}{4}} \cos(\ln(\tan(x)))dx$ and $\int_0^{\frac{\pi}{4}} \sin(\ln(\tan(x)))dx$

Question

I was trying to evaluate the integral

$$\ I = \int_0^{\frac{\pi}{4}} \cos(\ln(\tan(x)) dx$$

So I substitute $tanx = t$ and the integral becomes

$$\ I = \int_0^1 \cos(\ln(t))\cdot \frac{1}{1+t^2} dt$$

Then I let

$$\ J = \int_0^1 \sin(\ln(t))\cdot \frac{1}{1+t^2}dt$$

And it follows from this previous answer that:

$$\ I + iJ = \int_0^1 \frac{e^{i\operatorname{ln}t}}{1+t^2} dt = \int_0^1 \frac{t^{i}}{1+t^2} dt$$

$$ \ I + iJ = \frac{\pi}{2} \cdot \csc(\frac{\pi(i+1)}{2}) = \frac{\pi}{2}\cdot \sec(\frac{i\pi}{2}) = \frac{\pi}{2} \cdot \operatorname{sech}(\frac{\pi}{2}) $$

Does this imply

$$\int_0^1 \frac{\sin(\ln(t))}{1+t^2} = \int_0^{\frac{\pi}{4}} \sin(\ln(\tan(x))) dx = 0?$$

Because WolframAlpha is returning a different value

Can anyone help me verify whether the values for the integrals I have obtained are accurate or not?

Thank you for reading! Happy New Year $2024$ in advance!

Answer 1

As @RandomVariable pointed out, I got the upper bounds wrong.

$$\int_0^{\frac{\pi}{2}} \cos(\ln(\operatorname{tan}x)) dx = \frac{\pi}{2} \cdot \operatorname{sech}(\frac{\pi}{2})$$

I had computed the answer for $\frac{\pi}{2}$ instead of $\frac{\pi}{4}$.

From this, it is clear how to evaluate

$$\int_0^{1} \frac{\cos(\ln(t))}{t^2+1} dx$$

A common trick in such cases is to let

$v = \frac{1}{t}$

We then get

$$\int_0^1 \frac{\cos(\operatorname{ln}t)}{(1+t^2)}dt = \int_1^{\infty} \frac{\cos(\operatorname{ln}v)}{1+v^2} dv$$

Thus

$$ 2 I = \frac{\pi}{2}\cdot \operatorname{sech}(\frac{\pi}{2})$$

And we have our result

$$ I = \frac{\pi}{4} \cdot \operatorname{sech}(\frac{\pi}{2})$$

Note that the other integral

$$\int_0^{\frac{\pi}{2}} \sin(\ln(\operatorname{tan}x))dx = 0$$

By the same method we proved that

$$2\cdot \int_0^{\frac{\pi}{4}} \cos(\ln(\operatorname{tan}x))dx = \int_0^{\frac{\pi}{2}} \cos(\ln(\operatorname{tan}x)dx$$

Answer 2

A different method:

Using the known integral $$ I(p) = \int_{0}^{\infty} \frac{t^p \, dt}{1 + t^2} = \frac{\pi}{2} \, \sec\left(\frac{\pi \, p}{2}\right) $$ then \begin{align*} I(i) &= \int_{0}^{\infty} \frac{t^i \, dt}{1 + t^2} = \frac{\pi}{2} \,\operatorname{sech}\left(\frac{\pi}{2}\right) \\ I(-i) &= \int_{0}^{\infty} \frac{t^{-i} \, dt}{1 + t^2} = \frac{\pi}{2} \, \operatorname{sech}\left(\frac{\pi}{2}\right) \end{align*} and \begin{align*} \int_{0}^{\infty} \frac{t^i + t^{-i}}{1 + t^2} \, dt &= \pi \, \operatorname{sech}\left(\frac{\pi}{2}\right) \\ \int_{0}^{\infty} \frac{t^i - t^{-i}}{1 + t^2} \, dt &= 0. \end{align*}

Since $t^i = e^{i \, \ln(t)}$ then the last two integrals become \begin{align*} \int_{0}^{\infty} \frac{\cos(\ln t)}{1 + t^2} \, dt &= \frac{\pi}{2} \, \operatorname{sech}\left(\frac{\pi}{2}\right) \\ \int_{0}^{\infty} \frac{\sin(\ln t)}{1 + t^2} \, dt &= 0. \end{align*} Letting $ t = \tan(x) $ gives the integral results \begin{align*} \int_{0}^{\pi/2} \cos(\ln \tan x) \, dx &= \frac{\pi}{2} \, \operatorname{sech}\left(\frac{\pi}{2}\right) \\ \int_{0}^{\pi/2} \sin(\ln \tan x) \, dx &= 0. \end{align*}

Note that since $$ \int_{0}^{\pi/4} \cos(\ln(\tan x)) \, dx = \frac{1}{2} \, \int_{0}^{\pi/2} \cos(\ln(\tan x)) \, dx $$ then $$ \int_{0}^{\pi/4} \cos(\ln \tan x) \, dx = \frac{\pi}{4} \, \operatorname{sech}\left(\frac{\pi}{2}\right). $$

Note 2:

Using the known integral $$ \int_{0}^{1} \frac{t^p \, dt}{1 + t^2} = \frac{1}{4} \, \left(\psi\left(\frac{p+3}{4}\right) - \psi\left(\frac{p+1}{4}\right) \right). $$ then following the same pattern $$ \int_{0}^{\pi/4} \sin(\ln \tan x) \, dx = \frac{i}{8} \, \left(\psi\left(\frac{3-i}{4}\right) - \psi\left(\frac{3+i}{4}\right) + \psi\left(\frac{1+i}{4}\right) - \psi\left(\frac{1-i}{4}\right) \right). $$