# question about improper integral

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• Nov 27th 2008, 11:49 AM
gilgames
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
• Nov 27th 2008, 12:16 PM
TheEmptySet
Quote:

Originally Posted by gilgames
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
• Nov 27th 2008, 12:25 PM
gilgames
thanks for the response. i had tried the partial integration but could not find the result. thanks