Integral-serie

• Jan 2nd 2010, 09:33 AM
dhiab
Integral-serie
Calculate and student this integral :
$\displaystyle \int_0^{ + \infty } {\left( {\frac{{\arctan \left( {\frac{1}{x}} \right)}}{{x^2 + 1}}} \right)} dx$
Same question :
$\displaystyle \int_0^{ + \infty } {\left( {\sin \left( {\frac{1}{{x^\alpha }}} \right)} \right)} dx.....\left( {\alpha \in \Re } \right)$
• Jan 2nd 2010, 09:39 AM
pomp
Quote:

Originally Posted by dhiab
Calculate and student this integral :
$\displaystyle \int_0^{ + \infty } {\left( {\frac{{\arctan \left( {\frac{1}{x}} \right)}}{{x^2 + 1}}} \right)} dx$
Same question :
$\displaystyle \int_0^{ + \infty } {\left( {\sin \left( {\frac{1}{{x^\alpha }}} \right)} \right)} dx.....\left( {\alpha \in \Re } \right)$

You haven't shown any working, you rarely do. I can never tell if you're asking for help or posing a question which you know the answer to.

If you are seeking help then you should really show what working you have done and where you get stuck.

Here's a hint for the first one

$\displaystyle \frac{d}{dx} \left(\arctan (1/x) \right) = -\frac{1}{1+x^2}$

Hope this helps.
• Jan 2nd 2010, 11:45 AM
Krizalid
As for the second question, the integral obviously diverges for $\displaystyle \alpha\le0,$ so the only interesting case to analyze convergence is $\displaystyle \alpha>0.$ Here's a Hint: the integral converges for all $\displaystyle \alpha>1.$ Prove it.
• Jan 2nd 2010, 01:18 PM
Drexel28
Quote:

Originally Posted by Krizalid
As for the second question, the integral obviously diverges for $\displaystyle \alpha\le0,$ so the only interesting case to analyze convergence is $\displaystyle \alpha>0.$ Here's a Hint: the integral converges for all $\displaystyle \alpha>\frac12.$ Prove it.

Doesn't then integral converges for all $\displaystyle \alpha>1$? $\displaystyle \lim_{x\to\infty}\frac{\sin\left(\tfrac{1}{x^{\alp ha}}\right)}{\tfrac{1}{x^{\alpha}}}=1$. So that we can see that $\displaystyle \int_1^{\infty}\sin\left(\tfrac{1}{x^{\alpha}}\rig ht)dx$ shares convergence with $\displaystyle \int_1^{\infty}\frac{dx}{x^{\alpha}}$

[Yes, it was a typo, it's fixed now. -K.]