Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / D / D4

A sufficient differentiation theorem is the following local version. Suppose $u(t,\cdot )$ is integrable, $u(\cdot ,x)$ is continuously differentiable for almost every $x$, and for each compact parameter interval there is an integrable $g(x)$ bounding $|{\partial }_{t}u(t,x)|$ uniformly there. Then $I^{\prime }(t)={\int }_E{\partial }_{t}u(t,x)dx$. The mean value theorem bounds the difference quotients by $g$, and dominated convergence proves the formula. Integrability at one parameter value together with the derivative bound also suffices locally.

The $k$th derivative of the integrand is $(\cos x{)}^k\cos (t\cos x+k\pi /2)$, whose absolute value is at most one. The integration interval has finite measure. Repeated application of the differentiation theorem therefore givesDominated convergence also gives continuity of every derivative, proving infinite differentiability.

From (a), combine the three derivative integrals:The integrand is $-\frac{d}{dx}[\sin x\sin (t\cos x)]$. This bracket vanishes at both endpoints, so the integral is zero. Hence $\boxed{t{J}_0^{\prime \prime }+J_0^{\prime }+t{J}_0=0}$, also at $t=0$ by the same formula.

A direct derivative bound by $1/x$ would not be integrable, so the earlier theorem cannot be applied that way. Instead, for $t>0$ make the legitimate substitution $s=tx$ in the absolutely convergent integral:The tail integral has continuous integrand away from zero and converges absolutely. The fundamental theorem of calculus therefore gives derivative $-\cos t/t^2$ with respect to its lower endpoint. The product rule yieldsThis establishes differentiability carefully without an invalid domination argument.