My Math Forum Integration over open domain or close domain

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 October 11th, 2008, 09:40 PM #1 Newbie   Joined: Oct 2008 Posts: 4 Thanks: 0 Integration over open domain or close domain Hi I tried to solve begin $\int_0^{k_o}\frac{1}{ ( k-{k_o})^2 }dx$ using variation change $u= k-k_o, du = dk$ So integration is transformed to $\int_{-k_{o}}^0 \frac{1}{u^2} du$ and its calculation is $\frac{1}{u} |_{-k_0}^0$ (Please tell me how can i write integration reult with upper and bottom bound like a textbook using Latex!) But its result $\frac{1}{0} + \frac{1}{k_o}$is unacceptable becuase $\frac{1}{0}$ is not defined in mathematic. I wandered and when i read question more precisely i found that given range is not $[0,k_o]$ but $(0,k_o)$
October 12th, 2008, 09:26 AM   #2
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Re: Integration over open domain or close domain

Quote:
 Originally Posted by good_phy I wandered and when i read question more precisely i found that given range is not $[0,k_o]$ but $(0,k_o)$
So then what is the limit as x goes to 0 of $\frac{1}{0}$? Add that to $k_o$. Of course, that limit is not finite... So the integral does not exist.

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