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Thread: More improper integrals

  1. #1
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    More improper integrals

    Evaluate

    $\displaystyle \int_0^1\frac{x^{2007}-x^{2006}}{\ln x}\,dx$

    $\displaystyle \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}$

    $\displaystyle \int_0^\infty\frac{dx}{(r^2+2rx\cos\theta+x^2)^{3/2}}$
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  2. #2
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    1)
    Let $\displaystyle t=\ln x$ then:
    $\displaystyle \int_{-\infty}^0 \frac{e^{2008t} - e^{2007t}}{t}dt$.
    Thus,
    $\displaystyle \int_{-\infty}^0 \int_0^{\infty} (e^{2008t}-e^{2007t})e^{-\mu t}~d\mu ~dt$
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  3. #3
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    Quote Originally Posted by liyi View Post
    Evaluate

    $\displaystyle \int_0^1\frac{x^{2007}-x^{2006}}{\ln x}\,dx$
    Hehe, I've got another solution for this one.

    $\displaystyle \begin{aligned}
    \int_0^1\frac{x^{2007}-x^{2006}}{\ln x}\,dx&=\int_0^1\int_{2006}^{2007}x^u\,du\,dx\\
    &=\int_{2006}^{2007}\int_0^1x^u\,dx\,du\\
    &=\int_{2006}^{2007}\frac{du}{u+1}\\
    &=\ln\frac{2008}{2007}.
    \end{aligned}$


    You can also set $\displaystyle u=-\ln x$ and apply Frullani from there.
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  4. #4
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    Quote Originally Posted by liyi View Post
    $\displaystyle \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}$
    $\displaystyle \frac{1}{\sqrt{(x-a)(b-x)}} = \frac{1}{\sqrt{-(x-a)(x-b)}}$.
    Let $\displaystyle t=x - \frac{a+b}{2}$ and we get:
    $\displaystyle \frac{1}{\sqrt{-\left( t - \frac{a-b}{2} \right)\left( t + \frac{a-b}{2} \right)}} = \frac{1}{\sqrt{\left( \frac{a-b}{2}\right)^2 - t^2}}$.
    Thus the integral becomes,
    $\displaystyle \int_{a-b}^{b-a} \frac{dt}{\sqrt{\left( \frac{a-b}{2}\right)^2 - t^2}} = \left. \frac{a-b}{2} \cdot \sin^{-1} \frac{2t}{a-b} \right|_{a-b}^{a+b}$
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  5. #5
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    Another attempt.

    Quote Originally Posted by liyi View Post
    Evaluate

    $\displaystyle \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}$
    Define $\displaystyle u=x-a,$

    $\displaystyle \int_a^b {\frac{{dx}}
    {{\sqrt {(x - a)(b - x)} }}} = \int_0^{b - a} {\frac{{du}}
    {{\sqrt {u(b - u - a)} }}} .$

    One more substitution according to $\displaystyle u = \frac{{(b - a)(1 + \kappa )}}
    {2},$

    $\displaystyle \int_a^b {\frac{{dx}}
    {{\sqrt {(x - a)(b - x)} }}} = \int_{ - 1}^1 {\frac{{d\kappa }}
    {{\sqrt {1 - \kappa ^2 } }}} = 2\arcsin 1 = \pi .$
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  6. #6
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    Quote Originally Posted by liyi View Post
    Evaluate

    $\displaystyle \int_0^\infty\frac{dx}{(r^2+2rx\cos\theta+x^2)^{3/2}}$
    Rewrite the integral as

    $\displaystyle
    \int_0^\infty {\frac{{dx}}
    {{\Big[ {\left( {x + r\cos \theta } \right)^2 + r^2 \sin ^2 \theta } \Big]\sqrt {\left( {x + r\cos \theta } \right)^2 + r^2 \sin ^2 \theta } }}} .
    $

    Now here's the trick, make the substitution $\displaystyle u = \frac{{x + r\cos \theta }}
    {{r\sin \theta }}.$
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