Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$. Give a simple asymptotic equivalent of $f(x)$ as $x$ tends to $-1$.
Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$.
Give a simple asymptotic equivalent of $f(x)$ as $x$ tends to $-1$.