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    Improper Integrals

    I have no idea where to start with improper integrals... I'm trying to solve the integral of 1/(x^2)-25 dx from 6 to infinity and I have no clue how to begin. Also, how do I recognize if it is divergent?
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    Quote Originally Posted by tom ato View Post
    I have no idea where to start with improper integrals... I'm trying to solve the integral of 1/(x^2)-25 dx from 6 to infinity and I have no clue how to begin. Also, how do I recognize if it is divergent?
    To get you started.
    \int_6^b {\frac{{dx}}<br />
{{x^2  - 25}}}  = \frac{1}<br />
{{10}}\ln \left( {\frac{{b - 5}}<br />
{{b + 5}}} \right) + \frac{{\ln (11)}}<br />
{{10}}.
    Now find the limit, \lim _{b \to \infty } \left( {\frac{1}<br />
{{10}}\ln \left( {\frac{{b - 5}}<br />
{{b + 5}}} \right) + \frac{{\ln (11)}}<br />
{{10}}} \right)
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  3. #3
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    Since the integrand is positive for 6\le x<\infty, then the limit comparison test applies with \int_6^\infty\frac{dx}{x^2} and your integral converges.

    Now, to compute its value, note that \frac{1}{x^{2}-25}=\frac{1}{(x+5)(x-5)}=\frac{1}{10}\cdot \frac{(x+5)-(x-5)}{(x+5)(x-5)}=\frac{1}{10}\left( \frac{1}{x-5}-\frac{1}{x+5} \right).
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by tom ato View Post
    I have no idea where to start with improper integrals... I'm trying to solve the integral of 1/(x^2)-25 dx from 6 to infinity and I have no clue how to begin. Also, how do I recognize if it is divergent?
    First note that

    \begin{aligned}\frac{1}{x^2-25} & =\frac{1}{(x+5)(x-5)}\\ & =\frac{10}{10(x-5)(x+5)}\\ & =\frac{(x+5)-(x-5)}{10(x+5)(x-5)}\\ & = \frac{1}{10(x-5)}-\frac{1}{10(x+5)}\end{aligned}.

    Now note that \frac{1}{10}\int\frac{1}{x-5}-\frac{1}{x+5}\,dx=\frac{1}{10}\left(\ln\left|x-5\right|-\ln\left|x+5\right|\right)=\frac{1}{10}\ln\left|\f  rac{x-5}{x+5}\right|.

    Now when you evalute \int_6^{\infty}\frac{\,dx}{x^2-25}\,dx, its the same as \frac{1}{10}\int_6^{\infty}\frac{1}{x-5}-\frac{1}{x+5}\,dx=\frac{1}{10}\left.\left[\ln\left|\frac{x-5}{x+5}\right|\right]\right|_6^{\infty}.

    Spoiler:
    Now by the FTC, this is the same as \frac{1}{10}\left[\lim_{b\to\infty}\ln\left|\frac{b-5}{b+5}\right|\right]-\frac{1}{10}\ln\left|\frac{6-5}{6+5}\right|=\ln 1-\frac{1}{10}\ln\left(\frac{1}{11}\right)=\color{re  d}\boxed{\frac{1}{10}\ln 11}.


    Does this make sense?

    To answer the last question. If the integral is similar to \int_a^{\infty}\frac1{x^p}\,dx, and p\leq 1, then the integral is divergent. Also, the integral can be divergent if the integrand crosses a vertical asymptote.

    EDIT: Plato and Kriz beat me...
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