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anonymous

  • 5 years ago

calculate the integral

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  1. anonymous
    • 5 years ago
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    \[\int\limits_{-\infty}^{\infty}dz/z ^{2}+25\]

  2. anonymous
    • 5 years ago
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    So?

  3. TuringTest
    • 5 years ago
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    I'm working on it did you try a trig sub\[z=5\tan\theta\]?

  4. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    (1/a) arctan(x/a)+c

  6. anonymous
    • 5 years ago
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    a=5 over here

  7. anonymous
    • 5 years ago
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    so i guess it wld be (1/5)arctan(x/5)

  8. anonymous
    • 5 years ago
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    whoops replace the x with z

  9. TuringTest
    • 5 years ago
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    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
    • 5 years ago
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    well we plug in no?

  11. anonymous
    • 5 years ago
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    so tan ^(-1)(0)=0

  12. anonymous
    • 5 years ago
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    so we are left with (1/5)tan^(-1)(b/5)

  13. anonymous
    • 5 years ago
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    turing u showed that n was positive both times but the first one shldbe negative and second time should be positive no?

  14. TuringTest
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    k wait i have a question

  20. anonymous
    • 5 years ago
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    \[\int\limits_{-\infty}^{0}f(x) +\int\limits_{0}^{\infty}f(x) = 2\int\limits_{0}^{\infty}f(x)\]

  21. anonymous
    • 5 years ago
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    is this true?

  22. anonymous
    • 5 years ago
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    Turing are you there?

  23. TuringTest
    • 5 years ago
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    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
    • 5 years ago
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    ...just because the integrand is even this is true

  25. anonymous
    • 5 years ago
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    ohhhh thanks turing you r AWESOME :D

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