I need to find the value of this integral when $n$ goes to infinity:
$$\int_0^\infty \frac{\cos(nx)}{1+x^n}\,dx$$
It should have the same value regardless if it's a Lebesgue and Riemann integral, yet I've been stuck because the function does not converge simply in $[0,1]$, so I can't use the theorem of dominated convergence. I thought about using the TDC on $]1,+\infty[$, and use a majoration on $[0,1]$ but I can't find a majoration that would make the thing go to 0.
Any help would be appreciated, thanks in advance!
Best Answer
Using integration by part, one has \begin{eqnarray} &&\int_0^\infty \frac{\cos(nx)}{1+x^n}\,dx\\ &=&\frac1n\int_0^\infty \frac{1}{1+x^n}\,d\sin(nx)\\ &=&\frac{1}{1+x^n}\sin(nx)\bigg|_0^\infty+\int_0^\infty \frac{x^{n-1}\sin(nx)}{(1+x^n)^2}\,dx\\ &=&\int_0^\infty \frac{x^{n-1}\sin(nx)}{(1+x^n)^2}\,dx. \end{eqnarray} Note $$ \bigg|\frac{x^{n-1}\sin(nx)}{(1+x^n)^2}\bigg|\le g(x)$$ where $$ g(x)=\bigg\{\begin{array}{ll}x^{n-1}\text{ if }x\in[0,1],\\\frac{1}{1+x^n}\text{ if }x>1, \end{array} $$ and $$ \lim_{n\to\infty}\frac{x^{n-1}\sin(nx)}{(1+x^n)^2}=0 \text{ a.e.}, \int_0^\infty g(x)\;dx<\infty. $$ By the DCT, one has $$ \lim_{n\to\infty}\int_0^\infty \frac{\cos(nx)}{1+x^n}\,dx=\lim_{n\to\infty}\int_0^\infty \frac{x^{n-1}\sin(nx)}{(1+x^n)^2}\,dx=0. $$