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anonymous
 4 years ago
calculate the integral
anonymous
 4 years ago
calculate the integral

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{\infty}^{\infty}dz/z ^{2}+25\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I'm working on it did you try a trig sub\[z=5\tan\theta\]?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well we have a table ofontegrals at the back of the book and it shows how to find the diffeential. it is a spectial differential

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i guess it wld be (1/5)arctan(x/5)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whoops replace the x with z

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, ok forgot about that so now it's an improper integral and needs to be split up as far as I know\[\lim_{n \rightarrow \infty}\frac1 5\tan^{1}(\frac z5) _{n}^{0}+\lim_{n \rightarrow \infty}\frac1 5\tan^{1}(\frac z5) _{0}^{n}\]but I'm not sure how to do these limits, so there is probably another method... Who else has an idea?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so we are left with (1/5)tan^(1)(b/5)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0turing u showed that n was positive both times but the first one shldbe negative and second time should be positive no?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1no because they are different limits the first will go to infinity, the second to positive so n doesn't have to be positive or negative, it makes no difference\[\lim_{n\rightarrow \infty}\frac1 5\tan^{1}(\frac z5) _{n}^{0}+\lim_{n\rightarrow \infty}\frac1 5\tan^{1}(\frac z5) _{0}^{n}\]\[=\frac 15[\lim_{n\rightarrow \infty}\tan^{1}(\frac n5)+\lim_{n\rightarrow \infty}\tan^{1}(\frac n5)] \]now each limit is pi/2 so the answer will be \[\frac\pi 5\]but my problem is I get\[\frac1 5(\frac\pi 2+\frac\pi 2)=0\]somehow they are supposed to add though apparently, I must be missing something.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no one equation goes to negative infinity and the other to positive I think you solve the 2 parts of the equation sepaprately

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1the correct answer is pi/5, so it seems this is set up right I have done the limits separately, if that is what you mean. I just seem to be missing a negative sign somehwere

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok I will review it and try to see what went wrong Thanks:D I appreciate your help :D seriously i know it took long to type it out:D

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1anytime, sorry I am a bit preoccupied so I can't devote all my attention to finding the missing minus sign good luck!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0k wait i have a question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{\infty}^{0}f(x) +\int\limits_{0}^{\infty}f(x) = 2\int\limits_{0}^{\infty}f(x)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Turing are you there?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1sorry, hey I figure it out the integrand is even so we can rewrite this as\[2\int_{0}^{\infty}\frac{dz}{z^2+25}=\frac2 5\lim_{n \rightarrow \infty}\tan^{1}\frac n 5=\frac{\pi}{5}\]...oh but you just wrote that nice job :D

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1...just because the integrand is even this is true

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhhh thanks turing you r AWESOME :D
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