We prove the required differentiation under the integral sign directly. Fix , and take a compact parameter interval containing in its interior. The jointly continuous partial derivative is bounded on the compact rectangle by some .
For nonzero sufficiently close to zero, defineFor each , the mean value theorem in the parameter gives . The definition of the partial derivative also gives . Since is integrable on , dominated convergence applied along any sequence yieldsThusAt , repeat the argument on with . It proves that the right derivative exists and satisfies the same formula at zero. This is the appropriate meaning of differentiability on the closed half-line. In fact is continuous there: apply part (i) to the continuous function .
Remove the apparent singularity at by definingThis is jointly continuous: write using the unnormalized sinc function, whose value at zero is one. Its partial derivative in isalso at , since there . This partial derivative is jointly continuous on the whole strip.
The value at the single endpoint does not change a Lebesgue integral. Part (b), whose dominated convergence proof applies to , therefore makes continuously differentiable on , with a right derivative at zero, andFor , direct integration givesAt zero the same integral gives , so no division by zero is required. Also . The substitution rule identifies for , the sine integral.
Right derivative 2026-10-31
The right derivative is the displayed one-sided limit, when it exists as a finite number. It describes differentiation at the left endpoint of a closed interval without requiring the function to be defined to the left. An ordinary derivative at an interior point requires matching left and right derivatives. For example, the absolute value function has right derivative one at zero and left derivative minus one, so it has no ordinary derivative there.