 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ib/paper-4/15f/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

t > 0

and

γ > 0

f (t) = \frac{1}{2 πi} \int_{γ - i \infty}^{γ + i \infty} \frac{e ^{s t}}{s ^{2} ( s + 1 ) ^{2}} d s .

Close the upward vertical segment by a left semicircle of radius

R

centred at

γ

. On that arc

Re s \leq γ

, so

∣ e^{s t} ∣ \leq e^{γ t}

, while

∣ s^{2} (s + 1)^{2} ∣

is bounded below by a constant times

R^{4}

for large

R

. Its length is

π R

, hence the arc integral tends to zero as

O (R^{- 3})

. The contour is positively oriented and encloses the double poles at

0

and

- 1

Their residues are

Res_{s = 0} \frac{e ^{s t}}{s ^{2} ( s + 1 ) ^{2}} = \frac{d}{d s} \frac{e ^{s t}}{( s + 1 ) ^{2}}_{s = 0} = t - 2,

and

Res_{s = - 1} \frac{e ^{s t}}{s ^{2} ( s + 1 ) ^{2}} = \frac{d}{d s} \frac{e ^{s t}}{s ^{2}}_{s = - 1} = (t + 2) e^{- t} .

The residue theorem therefore yields

f (t) = t - 2 + (t + 2) e^{- t} (t \geq 0) .

The value at zero follows by continuity and is

f (0) = 0

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