Integral limit $\lim_{n \to \infty} \int_{\mathbb{R}^{+}} \frac{\sin(x/n)}{x^2+x} \; dx$ I need to compute the (Lebesgue) integral limit
$$\lim_{n \to \infty} \int_{\mathbb{R}^{+}} \frac{\sin(x/n)}{x^2+x} \mathrm dx$$
I think I need to use the dominated conergence theorem or a similar theorem, but I'm not sure how to proceed.
Note on $[1, \infty)$ the application of DCT is obvious, but the singularity at $0$ has me confused.
 A: There actually is no singularity: $$\lim_{x \to 0}\frac{\sin\left(\frac{x}{n}\right)}{x^2+x} = \frac{1}{n}$$
And since $\sin(x) < x$ for $x > 0$, $$0<\int_0^1 \frac{\sin\left(\frac{x}{n}\right)}{x^2+x} dx \le \int_0^1 \frac{\frac{x}{n}}{x^2+x} dx = \frac{1}{n}\int_0^1 \frac{1}{x+1} dx = \frac{\ln(2)}{n}$$
Since both sides converge to $0$, the integral also converges to $0$. I ignored $x > 1$ partly because you said you got it already, but mainly because I'm not great with applying the convergence theorems.
A: You can compute the exact value of the integral
$$\frac{\sin \left(\frac{x}{n}\right)}{x^2+x}=\frac{\sin \left(\frac{x}{n}\right)}{x}-\frac{\sin \left(\frac{x}{n}\right)}{x+1}$$
$$I(a)=\int\frac{\sin \left(\frac{x}{n}\right)}{x+a}\,dx=\cos \left(\frac{a}{n}\right) \text{Si}\left(\frac{x+a}{n}\right)-\sin
   \left(\frac{a}{n}\right) \text{Ci}\left(\frac{x+a}{n}\right)$$
$$J(a)=\int_0^\infty\frac{\sin \left(\frac{x}{n}\right)}{x+a}\,dx=\text{Ci}\left(\frac{a}{n}\right) \sin \left(\frac{a}{n}\right)+\frac{1}{2}
   \left(\pi -2 \text{Si}\left(\frac{a}{n}\right)\right) \cos
   \left(\frac{a}{n}\right)$$
$$K=\int_0^\infty\frac{\sin \left(\frac{x}{n}\right)}{x^2+x}=J(0)-J(1)$$
$$K=\frac{1}{2} \left(\pi-2 \text{Ci}\left(\frac{1}{n}\right) \sin
   \left(\frac{1}{n}\right)-\left(\pi -2 \text{Si}\left(\frac{1}{n}\right)\right)
   \cos \left(\frac{1}{n}\right) \right)$$ Now, using asymptotics
$$K=\frac{\log (n)+1-\gamma }{n}+\frac{\pi }{4 n^2}+O\left(\frac{1}{n^3}\right)$$ which already a good approximation for small values of $n$.
For example, for $n=4$, the exact value is $0.494246$ while the above approximation gives $0.501357$.
