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Math Help - definite integral #2

  1. #1
    Super Member Random Variable's Avatar
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    definite integral #2

     \displaystyle \int^{3}_{0} \frac{\big(x^{3}(3-x)\big)^{\frac{1}{4}}}{5-x} \ dx

    There is a worked solution on Wikipedia using contour integration. I find it quite confusing.

    Methods of contour integration - Wikipedia, the free encyclopedia

    However, you don't have to resort to contour integration to solve the problem. There's a much easier (albeit longer) way.
    Last edited by Random Variable; January 13th 2011 at 11:12 PM.
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  2. #2
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    Is this what you're after?

    Let x=\frac{3}{1+t^4}, then the integral equals:

    \displaystyle{<br />
18\int_{-\infty}^\infty \frac{t^4}{(t^4+1)^2(5t^4+2)} dt<br />
}

    and we know this integral is equal to 2\pi i times the sum of residues of the integrand in the upper half of the complex plane, which is simple, technical and tedious to calculate.

    Alternatively, this can be calculated with partial fractions, with the help of the following identity:

    \displaystyle{<br />
\int_0^\infty \frac{1}{1+x^a}dx = \frac{\pi}{a}\csc\frac{\pi}{a}<br />
}
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  3. #3
    Super Member Random Variable's Avatar
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    Is this what you're after?
    Not exactly.


    Rewrite the integral as  \displaystyle \int^{3}_{0} \frac{x}{5-x} \Big(\frac{3-x}{x}\Big)^{\frac{1}{4}}

    let  t = \frac{3-x}{x}

    then the integral becomes   \displaystyle 9 \int_{0}^{\infty} \frac{1}{(5t+2)(t+1)^{2}} \ t^{\frac{1}{4}} \ dt

    use partial fractions

     \displaystyle 25 \int^{\infty}_{0} \frac{t^{\frac{1}{4}}}{5t+2} \ dt t - 5 \int^{\infty}_{0} \frac{t^{\frac{1}{4}}}{t+1} \ dt - 3 \int^{\infty}_{0} \frac{t^{\frac{1}{4}}}{(t+1)^{2}} \ dt

    Now write each integral as a double integral.

     = \displaystyle 25 \int^{\infty}_{0} \int^{\infty}_{0} t^{\frac{1}{4}} \  e^{-(5t+2)y} \ dy \ dt - 5 \int^{\infty}_{0} \int^{\infty}_{0} t^{\frac{1}{4}} \ e^{-(t+1)y} \ dy \ dt- 3 \int^{\infty}_{0} \int^{\infty}_{0} \ t^{\frac{1}{4}} \ y \ e^{-(t+1)y} \ dy \ dt

    From here swtich the order of integration, integrate, and then use Euler's reflection formula and the property  \Gamma(z+1) = z \Gamma(z) .

    After some final manipulation/simplification, you'll get the answer.
    Last edited by Random Variable; January 14th 2011 at 10:34 AM.
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  4. #4
    Super Member Random Variable's Avatar
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    Wait a minute. The first two integrals after using partial fractions diverge. Is it even legal to do what I did next? How did I I get the correct answer?
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  5. #5
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    Quote Originally Posted by Random Variable View Post
    Wait a minute. The first two integrals after using partial fractions diverge. Is it even legal to do what I did next? How did I I get the correct answer?
    Yes, but summing the integrands of those two gives an integral which does converge.

    \displaystyle 5 \int^{\infty}_{0} t^{\frac{1}{4}}\left( \frac{5}{5t+2} - \frac{1}{t+1}\right) \ dt

    \displaystyle =5 \int^{\infty}_{0} \frac{3t^{\frac{1}{4}}}{(5t+2)(t+1)} \ dt
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  6. #6
    Super Member Random Variable's Avatar
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    It's valid to switch the order of integration of the first two double integrals since the integrands are always positive. The issue is what exactly is  \Gamma\Big(\frac{5}{4}\Big)\int^{\infty}_{0} y^{-5/4}e^{-y} \ dy . Is it infinity because the integral diverges? Or is it  \Gamma\Big(\frac{5}{4}\Big) \Gamma\Big(\frac{-1}{4} \Big) = \Gamma\Big(\frac{5}{4} \Big)\Gamma\Big(1-\frac{5}{4} \Big) = \frac{\pi}{\sin \frac{5 \pi}{4}}} = - \sqrt{2} \pi ?
    Last edited by Random Variable; January 14th 2011 at 01:40 PM.
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