For fixed $y$,
$$
\gamma(x,y)=\Gamma(x)-\int_0^y t^{x-1}e^{-t}\,dt.
$$
The omitted integral is at most $y^x/x$, which is negligible compared with $\Gamma(x)$ as $x\to\infty$. Stirling's formula therefore gives
$$
\boxed{
\gamma(x,y)\sim\Gamma(x)
\sim\sqrt{2\pi}\,x^{x-1/2}e^{-x}.}
$$

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 32D / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/32d/b/solution