Evaluate the integral ∫₀^∞ ln(x)/(x² + 1) dx using complex analysis techniques.

Step 1: Define the complex integral and contour

We need to evaluate the given definite integral. We will use complex analysis, specifically the residue theorem, to solve this. We consider a complex function $f(z) = \frac{\ln(z)}{z^2 + 1}$ and integrate it over a suitable contour.

Step 2: Identify singularities

The singularities of the function $f(z) = \frac{\ln(z)}{z^2 + 1}$ occur where the denominator is zero. Solving $z^2 + 1 = 0$ gives $z = i$ and $z = -i$. These are simple poles. The branch cut for $\ln(z)$ is typically chosen along the positive real axis, but for this integral, we can use a keyhole contour or a semicircular contour with a branch cut along the positive real axis.

Step 3: Choose contour and apply Residue Theorem

We choose a semicircular contour in the upper half-plane, indented at the origin, with a branch cut along the positive real axis. The only pole inside this contour is $z = i$. We calculate the residue at this pole.

Step 4: Calculate the residue at $z=i$

To find the residue at the simple pole $z=i$, we use the formula $\text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z)$. Substituting $z_0 = i$, we get $\frac{\ln(i)}{2i}$. We know that $\ln(i) = \ln(e^{i\pi/2}) = i\pi/2$.

Step 5: Substitute $\ln(i)$ and simplify residue

Substituting $\ln(i) = i\pi/2$ into the residue calculation, the $i$ terms cancel out, leaving us with $\frac{\pi}{4}$.

Step 6: Apply Residue Theorem to the contour integral

According to the Residue Theorem, the contour integral is $2\pi i$ times the sum of the residues. Since there is only one pole inside our chosen contour, the integral is $2\pi i \times \frac{\pi}{4} = \frac{i\pi^2}{2}$.

Step 7: Relate contour integral to the real integral

For the chosen contour, the integral along the large arc goes to zero as the radius approaches infinity, and the integral along the small arc around the origin goes to zero as the radius approaches zero. The integral along the path just above the positive real axis is $I$, and the integral along the path just below the positive real axis is also $I$ (due to the branch cut and the specific form of the integrand). Therefore, the total contour integral is $2I$. Equating this to the result from the Residue Theorem, we get $2I = \frac{i\pi^2}{2}$. However, this is incorrect. The integral along the real axis from 0 to $\infty$ is $I$. The integral along the real axis from $\infty$ to 0 (below the branch cut) is $\int_\infty^0 \frac{\ln(x) + 2\pi i}{x^2+1} dx = -\int_0^\infty \frac{\ln(x)}{x^2+1} dx - 2\pi i \int_0^\infty \frac{1}{x^2+1} dx = -I - 2\pi i [\arctan(x)]_0^\infty = -I - 2\pi i (\pi/2) = -I - i\pi^2$. Summing the contributions, we have $I + (-I - i\pi^2) = -i\pi^2$. This is also incorrect. Let's re-evaluate the contour. A common contour for this type of integral is a semi-circle in the upper half-plane. The integral along the real axis from $-R$ to $R$ is $\int_{-R}^R \frac{\ln(x)}{x^2+1} dx$. However, $\ln(x)$ is not defined for negative $x$. A better approach is to use the function $f(z) = \frac{\ln^2(z)}{z^2+1}$ or a keyhole contour. Let's use a keyhole contour for $f(z) = \frac{\ln(z)}{z^2+1}$. The integral along the upper edge of the branch cut is $I$. The integral along the lower edge of the branch cut is $\int_\infty^0 \frac{\ln(x) + 2\pi i}{x^2+1} dx = -I - 2\pi i \int_0^\infty \frac{1}{x^2+1} dx = -I - 2\pi i (\pi/2) = -I - i\pi^2$. The sum of these two parts is $I + (-I - i\pi^2) = -i\pi^2$. This should be equal to $2\pi i \text{Res}(f,i) = 2\pi i (\pi/4) = i\pi^2/2$. This implies $-i\pi^2 = i\pi^2/2$, which is false. Let's use a different approach. Consider the integral $J = \int_0^\infty \frac{\ln(x)}{x^2+1} dx$. Let $x = 1/t$, then $dx = -1/t^2 dt$. When $x=0$, $t=\infty$. When $x=\infty$, $t=0$. $J = \int_\infty^0 \frac{\ln(1/t)}{(1/t)^2+1} (-1/t^2) dt = \int_0^\infty \frac{-\ln(t)}{(1/t^2)+1} (1/t^2) dt = \int_0^\infty \frac{-\ln(t)}{1+t^2} dt = -J$. So $J = -J$, which implies $2J = 0$, so $J = 0$. This is a known result for this specific integral. Let's re-evaluate the complex analysis approach. Consider the integral $I = \int_0^\infty \frac{\ln(x)}{x^2+1} dx$. Let's use the contour integral of $f(z) = \frac{\ln(z)}{z^2+1}$ over a semi-circular contour in the upper half-plane, with the branch cut for $\ln(z)$ along the positive real axis. The integral along the real axis from $\epsilon$ to $R$ is $\int_\epsilon^R \frac{\ln(x)}{x^2+1} dx$. The integral along the large arc $C_R$ goes to 0 as $R \to \infty$. The integral along the small arc $C_\epsilon$ goes to 0 as $\epsilon \to 0$. The integral along the real axis from $-R$ to $-\epsilon$ is problematic due to $\ln(x)$. A standard method for this integral is to consider $f(z) = \frac{\ln(z)}{z^2+1}$ and use a keyhole contour. The integral along the upper side of the branch cut (from $\epsilon$ to $R$) is $\int_\epsilon^R \frac{\ln(x)}{x^2+1} dx$. The integral along the lower side of the branch cut (from $R$ to $\epsilon$) is $\int_R^\epsilon \frac{\ln(x) + 2\pi i}{x^2+1} dx = -\int_\epsilon^R \frac{\ln(x) + 2\pi i}{x^2+1} dx = -\int_\epsilon^R \frac{\ln(x)}{x^2+1} dx - 2\pi i \int_\epsilon^R \frac{1}{x^2+1} dx$. As $\epsilon \to 0$ and $R \to \infty$, the sum of these two integrals is: $I - (I + 2\pi i \int_0^\infty \frac{1}{x^2+1} dx) = -2\pi i \int_0^\infty \frac{1}{x^2+1} dx$. We know $\int_0^\infty \frac{1}{x^2+1} dx = [\arctan(x)]_0^\infty = \pi/2$. So, the sum of the integrals along the branch cut is $-2\pi i (\pi/2) = -i\pi^2$. By the Residue Theorem, this must be equal to $2\pi i \text{Res}(f, i) = 2\pi i (\pi/4) = i\pi^2/2$. So, $-i\pi^2 = i\pi^2/2$, which is still incorrect. Let's re-examine the substitution $x = 1/t$. $I = \int_0^\infty \frac{\ln(x)}{x^2+1} dx$. Let $x = \tan \theta$. Then $dx = \sec^2 \theta d\theta$. When $x=0$, $\theta=0$. When $x=\infty$, $\theta=\pi/2$. $I = \int_0^{\pi/2} \frac{\ln(\tan \theta)}{\tan^2 \theta + 1} \sec^2 \theta d\theta = \int_0^{\pi/2} \frac{\ln(\tan \theta)}{\sec^2 \theta} \sec^2 \theta d\theta = \int_0^{\pi/2} \ln(\tan \theta) d\theta$. This is a known integral. We know that $\int_0^{\pi/2} \ln(\sin x) dx = -\frac{\pi}{2} \ln(2)$ and $\int_0^{\pi/2} \ln(\cos x) dx = -\frac{\pi}{2} \ln(2)$. So, $\int_0^{\pi/2} \ln(\tan x) dx = \int_0^{\pi/2} (\ln(\sin x) - \ln(\cos x)) dx = \int_0^{\pi/2} \ln(\sin x) dx - \int_0^{\pi/2} \ln(\cos x) dx = -\frac{\pi}{2} \ln(2) - (-\frac{\pi}{2} \ln(2)) = 0$. So the integral is 0. The complex analysis approach for this integral is more subtle. Consider the integral $I = \int_0^\infty \frac{\ln(x)}{x^2+1} dx$. Let's use the contour integral of $f(z) = \frac{\ln(z)}{z^2+1}$ over a semi-circular contour in the upper half-plane. The integral along the real axis from $-R$ to $R$ is $\int_{-R}^R \frac{\ln(x)}{x^2+1} dx$. However, $\ln(x)$ is not well-defined for negative $x$ without a branch cut. If we use the principal branch of $\ln(z)$, then $\ln(x)$ for $x<0$ is $\ln(|x|) + i\pi$. So $\int_{-R}^0 \frac{\ln(x)}{x^2+1} dx = \int_{-R}^0 \frac{\ln(|x|) + i\pi}{x^2+1} dx$. Let $x = -t$. $\int_R^0 \frac{\ln(t) + i\pi}{t^2+1} (-dt) = \int_0^R \frac{\ln(t) + i\pi}{t^2+1} dt = \int_0^R \frac{\ln(t)}{t^2+1} dt + i\pi \int_0^R \frac{1}{t^2+1} dt$. So, $\oint_C f(z) dz = \int_0^R \frac{\ln(x)}{x^2+1} dx + \int_0^R \frac{\ln(x)}{x^2+1} dx + i\pi \int_0^R \frac{1}{x^2+1} dx + \int_{C_R} f(z) dz$. As $R \to \infty$, $\int_{C_R} f(z) dz \to 0$. So, $2I + i\pi \int_0^\infty \frac{1}{x^2+1} dx = 2\pi i \text{Res}(f, i)$. $2I + i\pi (\pi/2) = 2\pi i (\pi/4)$. $2I + i\pi^2/2 = i\pi^2/2$. $2I = 0 \implies I = 0$. This matches the previous result. Let's write down the steps for this approach. The contour is a large semi-circle in the upper half-plane, with radius $R$. The integral along the real axis is from $-R$ to $R$. For $x < 0$, $\ln(x) = \ln(|x|) + i\pi$. So $\int_{-R}^R \frac{\ln(x)}{x^2+1} dx = \int_{-R}^0 \frac{\ln(|x|) + i\pi}{x^2+1} dx + \int_0^R \frac{\ln(x)}{x^2+1} dx$. Let $x = -t$ in the first integral. $\int_R^0 \frac{\ln(t) + i\pi}{t^2+1} (-dt) = \int_0^R \frac{\ln(t)}{t^2+1} dt + i\pi \int_0^R \frac{1}{t^2+1} dt$. So, $\int_{-R}^R \frac{\ln(x)}{x^2+1} dx = \int_0^R \frac{\ln(x)}{x^2+1} dx + \int_0^R \frac{\ln(x)}{x^2+1} dx + i\pi \int_0^R \frac{1}{x^2+1} dx$. As $R \to \infty$, this becomes $2I + i\pi \int_0^\infty \frac{1}{x^2+1} dx = 2I + i\pi (\pi/2) = 2I + i\pi^2/2$. By the Residue Theorem, this is equal to $2\pi i \text{Res}(f, i) = 2\pi i (\pi/4) = i\pi^2/2$. Therefore, $2I + i\pi^2/2 = i\pi^2/2$. $2I = 0 \implies I = 0$.

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