This is a problem from Widder's Advanced Calculus (p. 9 chapter 11 $\S$1.4) I'm having trouble evaluating. Could I have a hint?
\begin{align}\lim_{x\rightarrow\infty}\left(\Gamma\left(1/x\right)\right)^{-1}\int_{0}^{x}\frac{|\sin\left(t\right)|}{t}\:dt&\overset{?}{=}\end{align}
I don't exactly know where to start.
On
Using $$ \int_0^x\frac{|\sin(t)|}t\text{d}t \approx \frac xn\sum_{k=1}^n\left|\text{sinc}\left(\frac {xk}n\right)\right| $$ we have $$ \begin{align} \lim_{x\to\infty}\frac{1}{\Gamma(1/x)}\int_0^x|\text{sinc}(t)|\text{d}t & = \lim_{x\to0}\frac{1}{\Gamma(x)}\int_0^{1/x}|\text{sinc}(t)|\text{d}t \\ & = \lim_{x\to0}\frac{1}{n}\sum_{k=1}^n\frac{1}{x\Gamma(x)}\left|\text{sinc}\left(\frac{k}{nx}\right)\right| \\ & = \lim_{x\to0}\sum_{k=1}^n\frac{\left|\sin\left(\frac{k}{nx}\right)\right|}{k\Gamma(x)} \\ & = \sum_{k=1}^n\lim_{x\to 0}\frac{\left|\sin\left(\frac{k}{nx}\right)\right|}{k\Gamma(x)} \\ & = 0 \end{align} $$ where we can evaluate the limit on the last line by noting that $\Gamma(x)\to\infty$ and $|\sin(x)| \le 1$.
On
Recall that the Gamma function has the series representation
$$\Gamma(z)=\frac1z -\gamma+\frac1{2!}\left(\gamma^2+\zeta(2)\right)z-\frac1{3!}\left(\gamma^3+3\gamma \zeta(2)+2\zeta(3)\right)z^2+O(z^3)$$
Then, $\lim_{x\to \infty}\Gamma(1/x)=\lim_{z\to 0}\Gamma(z)=\infty$
Also, we have
$$\Gamma'(z)=-\frac1{z^2} +\frac1{2!}\left(\gamma^2+\zeta(2)\right)-\frac2{3!}\left(\gamma^3+3\gamma \zeta(2)+2\zeta(3)\right)z+O(z^2)$$
Then, we can apply L'Hospital's Rule to the limit
$$\begin{align} \lim_{x\to \infty}\frac{\int_0^x \frac{|\sin t|}{t}\,dt}{\Gamma(1/x)}&=\lim_{z\to 0}\frac{\frac{|\sin (1/z)|}{(1/z)}\left(-\frac1{z^2}\right)}{\Gamma'(z)}\\\\ &=\lim_{z\to 0}\frac{-|\sin (1/z)|}{z\left(-\frac1{z^2}+O(1)\right)}\\\\ &=\lim_{z\to 0}\frac{z|\sin(1/z)|}{1+O(z)}\\\\ &=0 \end{align}$$
Hint:
$$\Gamma (1/x) \ge \int_0^1t^{1/x-1}e^{-t}\,dt \ge (1/e)\int_0^1t^{1/x-1}\,dt .$$