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

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

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.1you're asking if the integral is convergent or divergent?

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.1if \[\int_0^{\pi}\frac{d\theta}{\sqrt{\pi\theta}}\] is convergent, you can conclude that the integral in question is also convergent.

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

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.1why don't you evaluate the definite integral? if it is a value, then the integral is convergent.

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

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.1it 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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