# Find out if $\int _1^{\infty }\:\left(\frac{2+\sin^2\left(x+1\right)}{x^2e^x}\right)dx$ converges

Find out if $$\int _1^{\infty }\:\left(\frac{2+\sin^2\left(x+1\right)}{x^2e^x}\right)dx$$ converges. This integral is just unsolvable by me right now. Parts/sub nothing works for me. How can I test if it converges? The integral is so hard to evaluate.

• This function is positive on $[1,\infty),$ so bound the numerator by 3. The remaining integral is clearly finite. – RideTheWavelet Feb 10 '17 at 0:44

When asked about convergence or divergence of an integral, never try to evaluate it - it's hardly ever possible or useful. Instead, remember that (provided $f(x)$ is positive everywhere) if $g(x) > f(x)$ for all $x > a$ and $\int_a^\infty g(x)dx$ converges, then so does $\int_a^\infty f(x)dx$; conversely, if $g(x) > f(x)$ for all $x > a$ and $\int_a^\infty f(x)dx$ diverges, then $\int_a^\infty g(x)dx$ diverges as well. Both of these can be thought of in terms of area - in the first case, $\int_a^\infty f(x)dx$ is the area under a curve contained in the finite space $\int_a^\infty g(x)dx$, while in the second case $\int_a^\infty f(x)dx$ is an infinite space inside the area $\int_a^\infty g(x)dx$.
In this case, remember that $\sin^2(x + 1) < 1$ for all $x$, so $\frac{2 + \sin^2(x+1)}{x^2e^x} \leq \frac{3}{x^2e^x}$. Since $e^x > 1$ for all $x > 0$, $\frac{3}{x^2e^x} \leq \frac{3}{x^2}$. So the question becomes: Does $\int_1^\infty \frac{3}{x^2}dx$ converge? If so, then your integral does too.