integration sin(1/x)/x

[shakalandro]

How do I prove that the integral from 1 to infinity of sin(1/x)/x is convergent. I need to use either the integral comparison test or the integral ratio test. I can't seem to find the right comparison or analysis.

 

[galactus]

\(\displaystyle \int_{1}^{\infty}\frac{sin(\frac{1}{x})}{x}dx=\int_{0}^{1}\frac{sin(x)}{x}dx\)

\(\displaystyle \text{Sine Integral}=Si(x)=\int_{0}^{x}\frac{sin(x)}{x}dx\)

which converges and is known as the Sine Integral. Designated by Si(x). In this case, Si(1).

This can not be integrated by elementary methods. Therefore, make a comparison.

 

[shakalandro]

I know you need to make a comparison, the problem asks to make a comparison or use the integral ratio test. I just don't know what g(x) should be. g(x) being the function I compare sin(1/x)/x to.

 

[Deleted member 4993]

\(\displaystyle sin(\frac{1}{x}) \, \le \, \frac{1}{x}\)

 

[galactus]

Besides finding whether or not it converges, you can find what it converges to by using a series.

\(\displaystyle \frac{sin(x)}{x}=\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot x^{2n}}{(2n+1)!}\)

\(\displaystyle \int \sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot x^{2n}}{(2n+1)!}=\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot x^{2n+1}}{(2n+1)(2n+1)!}\)

\(\displaystyle =\int_{0}^{1}\left[x-\frac{x^{3}}{18}+\frac{x^{5}}{600}-\frac{x^{7}}{35280}+......\right]\approx 0.946...\)

 

[shakalandro]

\(\displaystyle sin(\frac{1}{x}) \, \le \, \frac{1}{x}\)

Thankyou, that's what I needed. This means that \(\displaystyle \frac{sin(1/x)}{x} \leq \frac{1}{x^2}\).