To solve the integral

$I = \int_{0}^{\infty} \frac{\ln(1+x^2)}{1+x^2} \, dx$

using contour integration and the given substitution:

$J = \int_{\text{contour}} \frac{\ln(i+z)}{1+z^2} \, dz,$

we will proceed as follows:

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Step 1: Analyze the integrand

The integrand $\frac{\ln(i+z)}{1+z^2}$ has poles at $z = \pm i$, where $1+z^2 = 0$. These are simple poles.

The branch point of $\ln(i+z)$ occurs at $z = -i$, so we must be careful to define the branch cut.

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Step 2: Define the contour

The contour consists of:

1. A line segment from $-R$ to $R$ along the real axis.
2. A semicircular arc above the real axis from $R$ to $-R$ in the upper half-plane.

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Step 3: Parametrize and compute $J$

The integral $J$ is expressed as:

$J = \int_{\text{line}} \frac{\ln(i+z)}{1+z^2} \, dz + \int_{\text{arc}} \frac{\ln(i+z)}{1+z^2} \, dz.$

Step 4: Contribution from the real axis

On the real axis, $\ln(i+z)$ can be expressed in terms of $\ln|i+z|$ and the argument. For $z = x \in \mathbb{R}$,

$\ln(i+x) = \ln|i+x| + i\arg(i+x),$

where $|i+x| = \sqrt{1+x^2}$ and $\arg(i+x) = \tan^{-1}(x)$.

Thus, the contribution along the real line is:

$\int_{-R}^R \frac{\ln(i+x)}{1+x^2} \, dx.$

Splitting into real and imaginary parts:

• The real part simplifies to terms involving $\ln(\sqrt{1+x^2}) = \frac{1}{2}\ln(1+x^2)$.
• The imaginary part involves terms of $\tan^{-1}(x)$, which will integrate to zero symmetrically about the origin.

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Step 5: Contribution from the arc

For the semicircular arc $|z| = R$ in the upper half-plane:

• As $R \to \infty$, $\ln(i+z) \sim \ln z$, and $\frac{1}{1+z^2} \sim \frac{1}{z^2}$.
• Thus, the integral on the arc vanishes as $R \to \infty$.

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Step 6: Residue at $z = i$

The residue at $z = i$ for the integrand $\frac{\ln(i+z)}{1+z^2}$ is:

$\text{Residue} = \lim_{z \to i} (z-i) \frac{\ln(i+z)}{1+z^2} = \frac{\ln(2i)}{2i}.$

Simplify $\ln(2i)$:

$\ln(2i) = \ln 2 + i\frac{\pi}{2}.$

Thus:

$\text{Residue} = \frac{\ln 2 + i\frac{\pi}{2}}{2i}.$

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Step 7: Combine contributions

Using the residue theorem:

$J = 2\pi i \cdot \text{Residue at } z = i.$

Compute:

$J = 2\pi i \cdot \frac{\ln 2 + i\frac{\pi}{2}}{2i} = \pi (\ln 2 - \frac{\pi}{2}i).$

Since the original integral $I$ corresponds to the real part of $J$, we have:

$I = \pi \ln 2.$

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Final Answer:

$\int_{0}^{\infty} \frac{\ln(1+x^2)}{1+x^2} \, dx = \pi \ln 2.$

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Would you like to explore details of any step or check for further simplifications? Here are 5 related questions:

1. What is the role of the residue theorem in contour integration?
2. How is the branch cut of $\ln(i+z)$ handled in this context?
3. Why does the integral along the semicircular arc vanish as $R \to \infty$?
4. Can this integral be solved using substitution instead of contour integration?
5. What is the relationship between the imaginary part of $J$ and the symmetry of the problem?

Tip: Always check the location of poles and branch points when setting up a contour for integration.