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anonymous
 3 years 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
anonymous
 3 years 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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits _{ 0 }^{ \pi }{ \frac { \sin\theta }{ \sqrt { \pi\quad \quad \theta } } } using\quad \frac { 1 }{ \sqrt { \pi\quad \quad \theta } } \quad as\quad comparison\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry idk how to do it, good luck though!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you're asking if the integral is convergent or divergent?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if \[\int_0^{\pi}\frac{d\theta}{\sqrt{\pi\theta}}\] is convergent, you can conclude that the integral in question is also convergent.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How do I know that integral you stated above is convergent? That's the part that is troubling me.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why don't you evaluate the definite integral? if it is a value, then the integral is convergent.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh okay, I thought it was improper. I just realized that it was definite. Thanks!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it is an improper integral, but after integration, the function is continuous in that interval, so you can treat it like an ordinary definite integral.
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