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