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

Asked Apr 15 '14 at 09:28

Active Aug 01 '16 at 12:23

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

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

edited Aug 01 '16 at 12:23

Yuriy S

31,474

asked Apr 15 '14 at 09:28

Shirly Geffen

2,571

Have you tried to estimate $\lfloor|x|\rfloor!$ using Stirling's formula? – 5xum Apr 15 '14 at 09:39
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

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

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