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Improper integral comparison test

  • Thread starter hitemup
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  • #1
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The question asks whether the following converges or diverges.

[tex]\int_{0}^{\infty } \frac{\left | sinx \right |}{x^2} dx[/tex]

Now I think there might be a trick with the domain of sine function but I couldn't make up my mind on this.
I tried to compare it with 1/x^2, (sinx)/x, and sinx. I actually expected that I would get something good with 1/x^2, but as the lower limit of the integral is zero, it ended up with infinity on (0, inf) and since 1/x^2 is greater than (sinx)/x^2, and is divergent as we just found, we cannot say whether the given function diverges or converges. So I'm wondering what is the right track on this question?
 
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Answers and Replies

  • #2
Stephen Tashi
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[tex]\int_{0}^{\infty } \frac{\left | sinx \right |}{x^2} dx = [/tex]

[tex] \int_{0}^{ \frac{\pi}{2} } \frac{\left | sinx \right |}{x^2} dx + \int_{ \frac{\pi}{2} }^{\infty} \frac{\left | sinx \right |}{x^2} dx [/tex]

Deal with the two integrals separately.

In the interval [itex] [0,\frac{\pi}{2}] [/itex] you can establish an inequality between [itex] |\sin(x)| [/itex] and a non-periodic function. Try comparing it to [itex] f(x) = x [/itex].
 

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