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Math Help - Finding a value for C, such that the integral Converges.

  1. #1
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    PLEASE HELP!!! Finding a value for C, such that the integral Converges.

    int(x/(x^2+1)-C/(3*x+1), x = 0 .. infinity)

    Please help me here. I thought that I would split the integral up into the two separate fractions and found that the first fraction diverged. I am unsure as to how I would go about finding the value of C.
    Please Help me.
    Last edited by nyhc_is_br00tal; March 31st 2008 at 07:48 PM.
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  2. #2
    Junior Member roy_zhang's Avatar
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    Quote Originally Posted by nyhc_is_br00tal View Post
    int(x/(x^2+1)-C/(3*x+1), x = 0 .. infinity)

    Please help me here. I thought that I would split the integral up into the two separate fractions and found that the first fraction diverged. I am unsure as to how I would go about finding the value of C.
    Please Help me.
    So the question is find value C such that \int\limits_0^\infty\left(\frac{x}{x^2+1}-\frac{C}{3x+1}\right)dx converges. Let's start to evaluate this improper integral as following:

    \int\limits_0^\infty\left(\frac{x}{x^2+1}-\frac{C}{3x+1}\right)dx

    =\lim\limits_{b\rightarrow \infty} \int\limits_0^b\left(\frac{x}{x^2+1}-\frac{C}{3x+1}\right)dx

    =\lim\limits_{b\rightarrow \infty}\left[\int\limits_0^b\left(\frac{x}{x^2+1}\right)dx-\int\limits_0^b\left(\frac{C}{3x+1}\right)dx\right]

    =\lim\limits_{b\rightarrow \infty}\left[\frac{1}{2}\ln(b^2+1)-\frac{C}{3}\ln(3b+1)\right]

    =\lim\limits_{b\rightarrow \infty}\left[\ln\frac{\sqrt{b^2+1}}{(3b+1)^\frac{C}{3}}\right]

    Now consider the expression \frac{\sqrt{b^2+1}}{(3b+1)^\frac{C}{3}}, since the highest power of b in the numerator is 1 (note that \sqrt{b^2}=b), we need the highest power of b in the denominator also to be 1 in order for the limit to converge. Otherwise, as b\rightarrow\infty, the expression \frac{\sqrt{b^2+1}}{(3b+1)^\frac{C}{3}} will either goto 0 or \infty, in either cases, the limit diverges. Hence when C=3, the given integral converges.


    Roy
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