Solution

Use the branch from part (b) and close the real interval

[- R, R]

with the upper semicircle, oriented counterclockwise. For

Im z \geq 0

z + i α

has positive imaginary part, so the numerator is a holomorphic logarithm throughout a neighbourhood of the contour and its interior. The branch point

- i α

lies strictly below the contour. The only enclosed pole of

H (z) = \frac{Log ( z + i α )}{z ^{2} + β ^{2}}

i β

Res_{z = i β} H = \frac{Log ( i ( α + β ))}{2 i β} = \frac{l o g ( α + β ) + iπ /2}{2 i β} .

To justify discarding the arc, take

R > max {β, α + 1}

. On it,

∣ z + i α ∣

lies between

R - α

and

R + α

, and the argument has bounded absolute value. Hence

\int_{arc} H (z) d z \leq \frac{π R { l o g ( R + α ) + 3 π /2 }}{R ^{2} - β ^{2}} ⟶ 0.

The real-axis integral is absolutely convergent, since its integrand is

O (lo g ∣ x ∣/ x^{2})

at infinity and has no real singularities. The residue theorem therefore gives

\int_{- \infty}^{\infty} \frac{Log ( x + i α )}{x ^{2} + β ^{2}} d x = \frac{π}{β} lo g (α + β) + \frac{i π ^{2}}{2 β} .

Finally,

Re Log (x + i α) = \frac{1}{2} lo g (x^{2} + α^{2})

. Taking twice the real part yields

\int_{- \infty}^{\infty} \frac{lo g ( x ^{2} + α ^{2} )}{x ^{2} + β ^{2}} d x = \frac{2 π}{β} lo g (α + β) .

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