# Improper Integrals

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• April 19th 2008, 07:40 PM
PaulRS
Quote:

Originally Posted by Mathstud28
ok $\int_0^{1}\frac{cos(x)}{x^2}dx=\int_0^{1}\sum_{n=1 }^{\infty}\frac{(-1)^{n}x^{2n-2}}{(2n)!}dx=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n-1}}{(2n-1)\cdot{(2n)!}}$ which converges from 0 to 1

But that's not the right series, did you read what TheEmptySet said?
• April 19th 2008, 07:43 PM
Mathstud28
Quote:

Originally Posted by PaulRS
But that's not the right series, did you read what TheEmptySet said?

Yes it is $\frac{cos(x)}{x^2}=\frac{\sum_{n=0}^{\infty}\frac{ (-1)^{n}x^{2n}}{(2n)!}}{x^2}=\sum_{n=1}^{\infty}\fra c{(-1)^{n}x^{2n-2}}{(2n)!}$
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