Can you generalize it?

$\displaystyle L =\lim_{x\to 0} \int_{ax}^{bx} \frac{\sin^nt}{t^{n+1}} dt$

Use the substitution: $u=\frac{t}{x} , du = \frac{1}{x}dt$ :

$\displaystyle L= \lim_{x\to 0} \int_{a}^{b} \frac{\sin^n(xu)}{(xu)^{n+1}} xdu$

$\displaystyle = \lim_{x\to 0} \int_{a}^{b} \frac{\sin^n(xu)}{(xu)^n} \frac{du}{u}$

$\displaystyle = \int_{a}^{b} \bigg(\lim_{x\to 0} \frac{\sin^n(xu)}{(xu)^n} \bigg) \frac{du}{u}$

$\displaystyle = \int_{a}^{b} (1^n) \frac{du}{u}$

$\displaystyle = \ln |u| |_{a}^{b} = \boxed{\ln |\frac{b}{a}| }$

For this problem, evaluate with $a= e , b=\pi$ to get the value $0.145$ to $3$ decimal places.