Answer: Integral of ln(1+x²)/(1+x²)

Key Idea / Intuition

The integrand $\frac{\ln(1+x^2)}{1+x^2}$ looks intimidating, but the substitution $x = \tan\theta$ turns $1+x^2$ into $\sec^2\theta$ and reveals a clean trigonometric integral. The $\ln(1+x^2)$ becomes $\ln(\sec^2\theta) = 2\ln(\sec\theta) = -2\ln(\cos\theta)$, and suddenly we're integrating $-2\ln(\cos\theta)$ over $[0,\pi/4]$, which can be handled by a classic symmetry/duplication trick.

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Formal Proof / Solution

Step 1: Substitution $x = \tan\theta$.

Let $x = \tan\theta$, so $dx = \sec^2\theta\, d\theta$ and $1+x^2 = \sec^2\theta$.

When $x=0$: $\theta=0$. When $x=1$: $\theta=\pi/4$.

$I = \int_0^{\pi/4} \frac{\ln(\sec^2\theta)}{\sec^2\theta}\cdot \sec^2\theta\, d\theta = \int_0^{\pi/4} \ln(\sec^2\theta)\, d\theta = -2\int_0^{\pi/4} \ln(\cos\theta)\, d\theta.$

Step 2: Evaluate $J = \int_0^{\pi/4} \ln(\cos\theta)\, d\theta$.

Use the known result (derivable via the Fourier series for $\ln(\cos\theta)$ or the standard Clausen integral):

$\int_0^{\pi/2} \ln(\cos\theta)\, d\theta = -\frac{\pi}{2}\ln 2.$

Split this into two pieces:

$\int_0^{\pi/2} \ln(\cos\theta)\, d\theta = \int_0^{\pi/4} \ln(\cos\theta)\, d\theta + \int_{\pi/4}^{\pi/2} \ln(\cos\theta)\, d\theta.$

In the second integral, substitute $\theta \to \frac{\pi}{2}-\phi$, so $\cos\theta = \sin\phi$:

$\int_{\pi/4}^{\pi/2} \ln(\cos\theta)\, d\theta = \int_0^{\pi/4} \ln(\sin\phi)\, d\phi.$

So:

$-\frac{\pi}{2}\ln 2 = \int_0^{\pi/4} \ln(\cos\theta)\, d\theta + \int_0^{\pi/4} \ln(\sin\theta)\, d\theta = \int_0^{\pi/4} \ln(\sin\theta\cos\theta)\, d\theta.$

Now use $\sin\theta\cos\theta = \frac{1}{2}\sin(2\theta)$:

$\int_0^{\pi/4} \ln\!\left(\tfrac{1}{2}\sin 2\theta\right)d\theta = \int_0^{\pi/4}\!\ln(\sin 2\theta)\,d\theta - \frac{\pi}{4}\ln 2.$

Substitute $u = 2\theta$ in the first part:

$\int_0^{\pi/4}\ln(\sin 2\theta)\,d\theta = \frac{1}{2}\int_0^{\pi/2}\ln(\sin u)\,du = \frac{1}{2}\cdot\left(-\frac{\pi}{2}\ln 2\right) = -\frac{\pi}{4}\ln 2.$

So:

$-\frac{\pi}{2}\ln 2 = -\frac{\pi}{4}\ln 2 - \frac{\pi}{4}\ln 2 = -\frac{\pi}{2}\ln 2. \checkmark$

This is consistent but doesn't separate $J$ directly. Instead, use the Clausen-type result directly:

$\int_0^{\pi/4}\ln(\cos\theta)\,d\theta = \frac{G}{2} - \frac{\pi}{4}\ln 2,$

where $G = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} \approx 0.9159...$ is Catalan's constant.

This follows from the Fourier series $-\ln(2\cos\theta) = \sum_{k=1}^\infty \frac{(-1)^k}{k}\cos(2k\theta)$, integrated term by term over $[0,\pi/4]$.

Step 3: Final answer.

$I = -2J = -2\left(\frac{G}{2} - \frac{\pi}{4}\ln 2\right) = -G + \frac{\pi}{2}\ln 2.$

$\boxed{I = \int_0^1 \frac{\ln(1+x^2)}{1+x^2}\,dx = \frac{\pi \ln 2}{2} - G}$

where $G$ is Catalan's constant. Numerically: $\frac{\pi\ln 2}{2} \approx 1.089$, $G \approx 0.916$, so $I \approx 0.173$.

The surprise: a clean-looking rational-times-log integrand over $[0,1]$ secretly encodes Catalan's constant — one of the most mysterious constants in mathematics, whose irrationality is still unknown.

Written to: questions/2025-07-14_PM_integral_ln_1_plus_x2.md