 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2010/d/d4/1/b/solution

Differentiating the integral representation gives

J_{0}^{'} (t) = - \frac{2}{π} \int_{0}^{π /2} cos x sin (t cos x) d x, J_{0}^{''} (t) + J_{0} (t) = \frac{2}{π} \int_{0}^{π /2} sin^{2} x cos (t cos x) d x .

Now integrate the derivative

\frac{d}{d x} (sin x sin (t cos x)) = cos x sin (t cos x) - t sin^{2} x cos (t cos x) .

The boundary terms vanish at both endpoints. Consequently

J_{0}^{'} (t) = - t (J_{0}^{''} (t) + J_{0} (t))

, proving

t J_{0}^{''} (t) + J_{0}^{'} (t) + t J_{0} (t) = 0.

The identity includes

t = 0

, where it reads

J_{0}^{'} (0) = 0

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