How do I compute the limit of this integral: $\lim_{n\to\infty}\int_{0}^{\infty}\frac{\sin(\frac{x}{n})}{(1+\frac{x}{n})^{n}}dx$?

Question

I'm trying to compute the limit of the integral $\displaystyle\int_{0}^{\infty}\frac{\sin(\frac{x}{n})}{(1+\frac{x}{n})^{n}}dx$ as $n$ goes to infinity. I tried and realized that this is probably not an elementary limit computation. I'm completely lost as to what result to use. Convergence theorems of Lebesgue integrals didn't get me far. Is this a problem of Fourier Analysis? I'd like to know of a way to do this.

Answer 1

Note that $\left(1+\dfrac{x}{n}\right)^{n}\geq 1+x+\dfrac{n(n-1)}{2}\dfrac{x^{2}}{n^{2}}\geq 1+\dfrac{1}{4}x^{2}$ whenever $n\geq 2$. So Lebesgue Dominated Convergence Theorem is applicable here.

The integrand is pointwise convergent to $0/e^{x}=0$. Now we see that $\left|\dfrac{\sin(x/n)}{(1+x/n)^{n}}\right|\leq\dfrac{1}{1+(1/4)x^{2}}$, as the later is $L^{1}(0,\infty)$ integrable, so the limit of the integral is zero.

Answer 2

What about using the Laplace transform?

$$ \int_{0}^{+\infty}\frac{\sin\tfrac{x}{n}}{\left(1+\tfrac{x}{n}\right)^n}\,dx = n\int_{0}^{+\infty}\frac{\sin x}{(x+1)^n}\,dx = \frac{n}{(n-1)!}\int_{0}^{+\infty}\frac{s^{n-1}\,ds}{(1+s^2)\,e^s}\tag{1}$$ and for every $n\geq 3$: $$ 0\leq \int_{0}^{+\infty}\frac{s^{n-1}\,ds}{(1+s^2)e^s}\leq \int_{0}^{+\infty}s^{n-3}e^{-s}\,ds = (n-3)! \tag{2}$$ hence: $$ \left| \int_{0}^{+\infty}\frac{\sin\tfrac{x}{n}}{\left(1+\tfrac{x}{n}\right)^n}\,dx \right|\leq \frac{n}{(n-1)(n-2)}\tag{3} $$ and the limit as $n\to +\infty$ is clearly zero.