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Math Help - question about improper integral

  1. #1
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    question about improper integral

    i could not find that integral converges or disconverges, please help me!

    the lower bounds of integral is 1 and the upper bound is infinite and f(x)=arctan(1/x^2). i am sorry that i don't know if there is any code to write it in mathematical expression
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  2. #2
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    Quote Originally Posted by gilgames View Post
    i could not find that integral converges or disconverges, please help me!

    the lower bounds of integral is 1 and the upper bound is infinite and f(x)=arctan(1/x^2). i am sorry that i don't know if there is any code to write it in mathematical expression
    \int_{1}^{\infty}\tan^{-1}\left( \frac{1}{x^2}\right)dx

    Integration by parts with u=\tan^{-1}\left( \frac{1}{x^2}\right) and dv=dx gives

    x\tan^{-1}\left( \frac{1}{x^2}\right)+\int_{1}^{\infty} \frac{2x^2}{x^4+1}dx < x\tan^{-1}\left( \frac{1}{x^2}\right)+\int_{1}^{\infty} \frac{2}{x^2}dx

    Since the right hand side converges by comparison the left hand side must also
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  3. #3
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    thanks for the response. i had tried the partial integration but could not find the result. thanks
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