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\int _{ 0 }^{ pi }{ \frac { sin\theta }{ \sqrt { pi\quad \quad \theta } } } using\quad \frac { 1 }{ \sqrt { pi\quad \quad \theta } } \quad as\quad comparison
 one year ago
 one year ago
\int _{ 0 }^{ pi }{ \frac { sin\theta }{ \sqrt { pi\quad \quad \theta } } } using\quad \frac { 1 }{ \sqrt { pi\quad \quad \theta } } \quad as\quad comparison
 one year ago
 one year ago

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David.ButlerBest ResponseYou've already chosen the best response.1
\[\int\limits _{ 0 }^{ \pi }{ \frac { \sin\theta }{ \sqrt { \pi\quad \quad \theta } } } using\quad \frac { 1 }{ \sqrt { \pi\quad \quad \theta } } \quad as\quad comparison\]
 one year ago

ChelseaSweetsBest ResponseYou've already chosen the best response.0
sorry idk how to do it, good luck though!
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
you're asking if the integral is convergent or divergent?
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
if \[\int_0^{\pi}\frac{d\theta}{\sqrt{\pi\theta}}\] is convergent, you can conclude that the integral in question is also convergent.
 one year ago

David.ButlerBest ResponseYou've already chosen the best response.1
How do I know that integral you stated above is convergent? That's the part that is troubling me.
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
why don't you evaluate the definite integral? if it is a value, then the integral is convergent.
 one year ago

David.ButlerBest ResponseYou've already chosen the best response.1
Oh okay, I thought it was improper. I just realized that it was definite. Thanks!
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
it is an improper integral, but after integration, the function is continuous in that interval, so you can treat it like an ordinary definite integral.
 one year ago
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