graça Escreveu:
\(\int_{-\infty}^{2} \frac{1}{x^{2}+4} dx\)
Olá
\(\int_{-\infty}^{2} \frac{1}{x^{2}+4} dx=\lim_{ p \rightarrow -\infty} \int_{p}^{2} \frac{1}{x^{2}+4}dx\)
calculando a integral definida : \(\int_{p}^{2} \frac{1}{x^{2}+4}\) vc obtém : \(\frac{arc tg(1)}{2}-\frac{arc tg (\frac{p}{2})}{2}\) , então:
\(\int_{-\infty}^{2} \frac{1}{x^{2}+4} dx=\lim_{ p \rightarrow -\infty} \frac{arc tg(1)}{2}-\frac{arc tg (\frac{p}{2})}{2}\)
\(\int_{-\infty}^{2} \frac{1}{x^{2}+4} dx=\lim_{ p \rightarrow -\infty} \frac{\pi}{8}-\frac{arc tg (\frac{p}{2})}{2}\)
\(\int_{-\infty}^{2} \frac{1}{x^{2}+4} dx= \frac{\pi}{8}+\frac{\pi}{4}\)
\(\int_{-\infty}^{2} \frac{1}{x^{2}+4} dx= \frac{3\pi}{8}\)
Então a integral converge,já que tende a um número real.