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watchmath
 5 years ago
Compute integral sin x /x from 0 to infinity.
watchmath
 5 years ago
Compute integral sin x /x from 0 to infinity.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to memorize the solutions to these. Don't try to actually solve it. The answer is just \pi/2\

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0I know the result. I want to know how :D.

M
 5 years ago
Best ResponseYou've already chosen the best response.0you'll have hard time getting help here for anything above precalculus

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you are currently in cal 1, you probably won't do the series stuff until the end of cal 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm going to upload an attachment soon, just hold on. Integration by parts SUCKS

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i think this involves advance calculus

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This does involve advanced calculus... this is going to take longer than I thought

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0can you tell me the idea?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you can use laplace transform

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0http://answers.yahoo.com/question/index?qid=20080129204110AAnGtsz

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is going to use Taylor Series convergence

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Honestly, I don't feel like proving that the Taylor Series converges. That would take about a week by hand for me (because I don't actually know how to do it yet). However, I'm assuming that it converges to pi/2

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0That's ok. I found it :D.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we can do this using complex analysis:\[\frac{1}{2}\int\limits_{\infty}^{\infty}\frac{\sin(x)}{x}dx=\frac{1}{2}I \left( \int\limits_{\infty}^{\infty}\frac{e^{ix}}{x}dx\right)\] this is the same as: \[I(\frac{1}{2} \pi i ( res_{0}\frac{e^{ix}}{x}))=I(\frac{\pi i}{2})=\frac{\pi}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we need the residue at zero since we have a simple pole there
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