## anonymous 5 years ago calculate the integral

1. anonymous

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

2. anonymous

So?

3. TuringTest

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

4. anonymous

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

5. anonymous

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

6. anonymous

a=5 over here

7. anonymous

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

8. anonymous

whoops replace the x with z

9. TuringTest

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?

10. anonymous

well we plug in no?

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. TuringTest

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.

15. anonymous

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

16. TuringTest

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

17. anonymous

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

18. TuringTest

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

19. anonymous

k wait i have a question

20. anonymous

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

21. anonymous

is this true?

22. anonymous

Turing are you there?

23. TuringTest

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

24. TuringTest

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

25. anonymous

ohhhh thanks turing you r AWESOME :D