 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/s1/2/4/solution

n = 3

- 6 πi I_{2} + 12 π^{2} I_{1} + 8 π^{3} i I_{0}

, while the residue result is

13 i π^{4} /4

. Substitution of

I_{0}, I_{1}

yields

I_{2} = π^{3} /8

n = 5

, the odd moments vanish and the remaining relation is

- 10 πi I_{4} + 80 π^{3} i I_{2} - 32 π^{5} i I_{0} = - \frac{121}{16} i π^{6} .

Thus

I_{2} = \frac{π ^{3}}{8}, I_{4} = \frac{5 π ^{5}}{32} .

As an independent generating-function check,

F (s) = \int_{0}^{\infty} x^{s} / (1 + x^{2}) d x = π / [2 cos (π s /2)]

for

∣ Re s ∣ < 1

. Differentiating twice and four times at zero gives the same moments.

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