关于积分∫₀^(π/4)cos(ln(tan(x)))dx与∫₀^(π/4)sin(ln(tan(x)))dx的求值正确性验证

Hey there! Let's break down where things went wrong in your derivation, and figure out the correct values for these integrals.

First off, your variable substitution steps are totally on point:

• Setting $\tan x = t$ correctly transforms $I = \int_0^{\frac{\pi}{4}} \cos(\ln(\tan x))dx$ into $I = \int_0^1 \frac{\cos(\ln t)}{1+t^2}dt$, and defining $J = \int_0^1 \frac{\sin(\ln t)}{1+t^2}dt$ to form the complex integral $I+iJ = \int_0^1 \frac{t^i}{1+t2}dt$ is a smart move—no issues here.

The Critical Mistake: Misapplying the Infinite Integral Formula #

Your error creeps in at the step where you wrote:
$$I + iJ = \frac{\pi}{2} \cdot \csc\left(\frac{\pi(i+1)}{2}\right)$$
This formula is only valid for the infinite integral $\int_0^\infty \frac{t^s}{1+t2}dt$, not the integral from 0 to 1. The infinite integral does equal $\frac{\pi}{2}\csc\left(\frac{\pi(s+1)}{2}\right)$ for $\text{Re}(s) \in (-1,1)$, but that's not what we're computing here.

Correct Approach to Compute $\int_0^1 \frac{t^i}{1+t2}dt$ #

To fix this, let's split the infinite integral into two parts:
$$\int_0^\infty \frac{t^i}{1+t2}dt = \int_0^1 \frac{t^i}{1+t2}dt + \int_1^\infty \frac{t^i}{1+t2}dt$$
For the second integral, substitute $u = 1/t$ (so $dt = -u^{-2}du$):
$$\int_1^\infty \frac{t^i}{1+t2}dt = \int_0^1 \frac{u^{-i}}{1+u2}du$$
Since the conjugate of $t^i$ is $t^{-i}$, we know $\int_0^1 \frac{t^{-i}}{1+t2}dt = I - iJ$. Now we can write:
$$\int_0^\infty \frac{t^i}{1+t2}dt = (I+iJ) + (I-iJ) = 2I$$
We already know the infinite integral's value:
$$\frac{\pi}{2}\csc\left(\frac{\pi(1+i)}{2}\right) = \frac{\pi}{2}\sec\left(\frac{i\pi}{2}\right) = \frac{\pi}{2}\text{sech}\left(\frac{\pi}{2}\right)$$
So solving for $I$ gives:
$$I = \frac{\pi}{4}\text{sech}\left(\frac{\pi}{2}\right)$$
Your original result for the cosine integral is actually correct! Now let's find $J$.

To calculate $J$, use the difference between the two integrals:
$$(I+iJ) - (I-iJ) = 2iJ = \int_0^1 \frac{t^i - t^{-i}}{1+t2}dt$$
We can expand $\frac{1}{1+t^2} = \sum_{n=0}^\infty (-1)^n t^{2n}$ (valid for $|t|<1$), so:
$$\int_0^1 \frac{t^i - t^{-i}}{1+t2}dt = \sum_{n=0}^\infty (-1)^n \left( \int_0^1 t^{2n+i}dt - \int_0^1 t^{2n-i}dt \right)$$
Each integral pair evaluates to:
$$\frac{1}{2n+i+1} - \frac{1}{2n-i+1} = \frac{-2i}{(2n+1)^2 + 1}$$
Substitute back and cancel $2i$ from both sides:
$$J = \sum_{n=0}^\infty (-1)^{n+1} \frac{1}{(2n+1)^2 + 1}$$
Calculating the first few terms gives us a non-zero value: $-0.5 + 0.1 - 0.03846 + 0.02 - ... ≈ -0.41846$, which matches WolframAlpha's output.

Summary #

• Your value for $I$ (the cosine integral) is accurate: $\int_0^{\frac{\pi}{4}} \cos(\ln(\tan x))dx = \frac{\pi}{4}\text{sech}\left(\frac{\pi}{2}\right)$
• The error was applying the infinite integral formula to the 0-to-1 integral, leading you to incorrectly conclude $J=0$. $J$ is actually a non-zero negative number, aligning with WolframAlpha's result.

Hope that clears things up! Happy New Year 2024!

备注：内容来源于stack exchange，提问作者user1173615