some help with contour integration?

Here's (1) for now; I'm still working on (2).

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1) Fix R > 1, and consider the usual semicircular contour C consisting of γ, the arc of radius R in the upper half plane, and the part of the real axis [-R, R].

So, ∫c log(z + i) dz/(z² + 1) = ∫γ log(z + i) dz/(z² + 1) + ∫(-R to R) log(x + i) dx/(x² + 1)

In C, the integrand log(z + i)/(z² + 1) only has a simple pole at z = i, with residue

lim(z→i) (z - i) * log(z + i)/(z² + 1) = lim(z→i) log(z + i)/(z + i) = log(2i)/(2i).

By the Residue Theorem,

∫c log(z + i) dz/(z² + 1) = 2πi * log(2i)/(2i) = π log(2i) = π ln 2 + iπ²/2.

As R→∞, |∫γ log(z+i) dz/(z² + 1)|→0 (essentially because log x = O(x^a) for any a > 0).

So, we obtain π ln 2 + iπ²/2 = 0 + ∫(-∞ to ∞) log(x + i) dx/(x² + 1).

==> ∫(-∞ to ∞) log(x + i) dx/(x² + 1) = π ln 2 + iπ²/2.

==> ∫(-∞ to ∞) [log√(x² + 1) + i Arg(x + i)] dx/(x² + 1) = π ln 2 + iπ²/2.

==> ∫(-∞ to ∞) log√(x² + 1) dx/(x² + 1) = π ln 2, equate real parts

==> 2 * ∫(x = 0 to ∞) (1/2) log(x² + 1) dx/(x² + 1) = π ln 2, integrand is even

==> ∫(x = 0 to ∞) log(x² + 1) dx/(x² + 1) = π ln 2.

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2) Let C be a 'keyhole' contour centered at z = 0, consisting of Cε (clockwise circle of radius ε about z = 0), γ (counterclockwise upper semicircle of radius R > 1), and segments on the real axis to and from x = ε to x = R.

**Note that arg will run from 0 to 2π as we go counterclockwise; this form of arg will be used in dealing with log z in this integral (instead of -π to π).

So, with f(z) = (log z)²/(z² + z + 1), we have

∫c f(z) dz = ∫γ (log z)² dz/(z² + z + 1) + ∫(R to ε) (log(x) + 2πi)² dx/(x² + x + 1)

+ ∫Cε (log z)² dz/(z² + z + 1) + ∫(ε to R) (log x)² dx/(x² + x + 1), due to the branch cut.

Note that f has singularities (simple poles again) when z² + z + 1 = 0.

==> z = (-1 ± i√3)/2 = e^(2πik/3), k = 1, 2.

Computing the residues (as before):

At z = e^(2πi/3): 4iπ²/(9√3).

At z = e^(4πi/3): -16iπ²/(9√3).

By the Residue Theorem, ∫c f(z) dz = 2πi [4iπ²/(9√3) - 16iπ²/(9√3)] = 8π³/(3√3).

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As before, lim(R→∞) |∫γ f(z) dz| = 0, and lim(ε→0+) ∫Cε f(z) dz = 0, by L'Hopital's Rule.

So, we obtain as R→∞ and ε→0+

8π³/(3√3) = 0 + ∫(∞ to 0) (log(x) + 2πi)² dx/(x² + x + 1) + 0 + ∫(0 to ∞) (log x)² dx/(x² + x + 1)

==> 8π³/(3√3) = ∫(x = 0 to ∞) (-4πi log(x) + 4π²) dx/(x² + x + 1)

Equate real and imaginary parts to obtain

∫(x = 0 to ∞) dx/(x² + x + 1) = 2π/(3√3) and ∫(x = 0 to ∞) log x dx/(x² + x + 1) = 0.

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I hope this helps!