# Convergence (absolutely) of an improper integral $\int_{-\infty}^\infty\frac{\sin(\sin x)}{1+\log(\lfloor|x|\rfloor! + 2)} dx$

$$\int_{-\infty}^\infty\frac{\sin(\sin x)}{1+\log(\lfloor|x|\rfloor! + 2)} dx$$

I need to check if this integral is absolutely convergent... I've shown it's convergent (not absolutely), according dirichlet test. I think it's conditionally convergent, but I don't know how to prove it.

Thank you

• Have you tried to estimate $\lfloor|x|\rfloor!$ using Stirling's formula? – 5xum Apr 15 '14 at 9:39
• I believe it is zero, since it is an odd function in a symmetric interval. – Mark Fantini Apr 15 '14 at 10:27
• @Fantini But then $\int_{-\infty}^{\infty} x dx = 0$ which is only true in a sense, $\lim_{n \to \infty} \int_{-n}^{n} x dx = 0$, but (what you would expect to be the same integral) $\lim_{n \to \infty} \int_{-n}^{2n} x dx$ does not exist – CameronJWhitehead Apr 15 '14 at 10:37
• @CameronJWhitehead You're right, it's zero in the Cauchy principal value sense. Guess we get no additional insights from it. – Mark Fantini Apr 15 '14 at 10:46
• @5xum I will try Stirling's formula, great idea. – Shirly Geffen Apr 15 '14 at 12:20