Question

calculate the integral

Answer 1

\[\int\limits_{-\infty}^{\infty}dz/z ^{2}+25\]

Answer 2

So?

Answer 3

I'm working on it
did you try a trig sub\[z=5\tan\theta\]?

Answer 4

Well we have a table ofontegrals at the back of the book and it shows how to find the diffeential. it is a spectial differential

Answer 5

(1/a) arctan(x/a)+c

Answer 6

a=5 over here

Answer 7

so i guess it wld be (1/5)arctan(x/5)

Answer 8

whoops replace the x with z

Answer 9

yeah, 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?

Answer 10

well we plug in no?

Answer 11

so tan ^(-1)(0)=0

Answer 12

so we are left with (1/5)tan^(-1)(b/5)

Answer 13

turing u showed that n was positive both times but the first one shldbe negative and second time should be positive no?

Answer 14

no 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.

Answer 15

no one equation goes to negative infinity and the other to positive I think you solve the 2 parts of the equation sepaprately

Answer 16

the 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

Answer 17

ok 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

Answer 18

anytime, sorry I am a bit preoccupied so I can't devote all my attention to finding the missing minus sign
good luck!

Answer 19

k wait i have a question

Answer 20

\[\int\limits_{-\infty}^{0}f(x) +\int\limits_{0}^{\infty}f(x) = 2\int\limits_{0}^{\infty}f(x)\]

Answer 21

is this true?

Answer 22

Turing are you there?

Answer 23

sorry, 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

Answer 24

...just because the integrand is even this is true

Answer 25

ohhhh thanks turing you r AWESOME :D