# Investigate absolute convergence of the integral $\int_0^\infty x^2\cos e^x\,dx$

I am studying absolute convergence of improper integral over $\left[0,+\infty\right)$ $$\int_0^\infty\!x^2\ \cos(e^x)\ dx$$ And I used the substitution $t=e^x$, I produce the improper integral $$\int_1^\infty\frac{(\ln\ t)^2\cos\ t}t\ dt$$ Thank you for your corrections

• Why would you use Dirichlet to study absolute convergence? – zhw. Oct 30 '15 at 5:17
• Because I don't know how the integral converge absolute while I could for conditionally convergence, that is my queation, how – Camilo Acevedo. Oct 30 '15 at 5:56
• See also Riemann-Lebesgue lemma. – Lucian Oct 30 '15 at 8:04
• The integral of the absolute value diverges. – zhw. Oct 30 '15 at 9:00
• @Lucian How could the RL lemma help? – zhw. Oct 30 '15 at 9:01

The highly oscillatory term $\cos(e^x)$, which makes the integral convergent in the improper sense, does not help with absolute convergence at all. A typical way to show this is to use the inequality $$|\cos t|\ge \cos^2 t = \frac12+\frac12\cos 2t$$ where the constant term $1/2$ is what kills convergence. The second term, with $\cos 2t$, produces an improperly convergent integral.
Precisely: $$\int_0^A x^2\cos e^x \,dx \ge \frac12 \int_0^A x^2\,dx + \frac12 \int_0^A x^2\cos 2 e^x \,dx$$ where the first term obviously grows and the second is
$$\frac12 \int_0^A x^2 e^{-x} ( e^x \cos 2 e^x) \,dx$$ where $x^2 e^{-x}$ decreases to zero (for large $x$) and the term in parentheses has a bounded antiderivative. Hence, the Dirichlet test applies to show that this has a limit as $A\to\infty$.