 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2017/s1/2/3/solution

Taking the difference of the two bank integrals at

n = 2

gives

\int_{0}^{\infty} \frac{( l o g x ) ^{2} - ( l o g x + 2 πi ) ^{2}}{1 + x ^{2}} d x = 4 π^{2} I_{0} - 4 πi I_{1} = 2 π^{3} .

The

I_{j}

are real and convergent. Comparing real and imaginary parts proves

I_{0} = π /2, I_{1} = 0

. Independently,

x \mapsto 1/ x

changes the sign of each odd logarithmic moment and preserves the measure, so every odd

I_{j}

vanishes.

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