 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/20/c/i/solution

Write

I (x) = ln g (x)

. For

x > - 1

I^{'} (x) = \int_{0}^{1} \frac{\partial}{\partial x} (\frac{t ^{x} - 1}{l n t}) d t = \int_{0}^{1} t^{x} d t = \frac{1}{x + 1} .

Indeed, on any compact parameter interval

c \leq x \leq d

with

c > - 1

, the differentiated integrand is bounded by the integrable function

t^{c}

. The dominated convergence theorem therefore justifies the operation despite the improper lower endpoint. The apparent singularity at

t = 1

is removable.

Since

g = e^{I}

g^{'} (x) = \frac{g ( x )}{x + 1}, x > - 1.

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