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Math Help - More improper integrals

  1. #1
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    More improper integrals

    Evaluate

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

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

    \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 t=\ln x then:
    \int_{-\infty}^0 \frac{e^{2008t} - e^{2007t}}{t}dt.
    Thus,
    \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

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

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


    You can also set 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
    \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}
    \frac{1}{\sqrt{(x-a)(b-x)}} = \frac{1}{\sqrt{-(x-a)(x-b)}}.
    Let t=x - \frac{a+b}{2} and we get:
    \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,
    \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

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

    \int_a^b {\frac{{dx}}<br />
{{\sqrt {(x - a)(b - x)} }}} = \int_0^{b - a} {\frac{{du}}<br />
{{\sqrt {u(b - u - a)} }}} .

    One more substitution according to u = \frac{{(b - a)(1 + \kappa )}}<br />
{2},

    \int_a^b {\frac{{dx}}<br />
{{\sqrt {(x - a)(b - x)} }}} = \int_{ - 1}^1 {\frac{{d\kappa }}<br />
{{\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

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

     <br />
\int_0^\infty {\frac{{dx}}<br />
{{\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 } }}} .<br />

    Now here's the trick, make the substitution u = \frac{{x + r\cos \theta }}<br />
{{r\sin \theta }}.
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