Proving a limit with Dominated convergence theorem

Question

For every $a>0$ let $$f_a(x):=\frac{1-e^{-ax}(\cos a+x\sin a)}{1+x^2},\qquad x\geq 0.$$

I have to prove that

$$\lim_{a\to +\infty }\int_{0}^{+\infty}f_a(x)dx=\int_{0}^{+\infty}\frac{1}{1+x^2}dx.$$

I think this can be done with the use of Dominated Convergence Theorem, since for every $x>0$ $$\lim_{a\to +\infty}f_a(x)=\frac{1}{1+x^2}.$$ The last thing I have to show is that for every $a$, $|f_a|\leq g$, with $g$ a nonnegative function such that $\int_{0}^{+\infty}g<+\infty$. Probably I can use $g(x)=\frac{2}{1+x^2}$, but I cannot prove that $|f_a|\leq g$ .

Any help would be really appreciated.

Answer 1

$|f_a(x)| \leq \frac {1+e^{-ax}(1+x)} {1+x^{2}} \leq \frac {1+e^{-x}(1+x)} {1+x^{2}}$ whenever $a>1$. Take $g(x)=\frac {1+e^{-x}(1+x)} {1+x^{2}}$.

$0 \leq e^{-x}(1+x)\leq 1$ on $[0,\infty)$ so $g$ is integrable.

Answer 2

By the Cauchy-Schwarz inequality $$ \left|e^{-ax}\frac{\cos a+x\sin a}{1+x^2}\right|\leq \frac{e^{-ax}}{\sqrt{1+x^2}}, $$ $$ \int_{0}^{+\infty}\frac{e^{-ax}}{\sqrt{1+x^2}}\,dx\leq\sqrt{\int_{0}^{+\infty}e^{-2ax}\,dx\int_{0}^{+\infty}\frac{dx}{1+x^2}}=\frac{1}{2}\sqrt{\frac{\pi}{a}}$$ so you do not really need the DCT.