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Math Help - Advanced Integration Techniques Textbook?

  1. #1
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    Advanced Integration Techniques Textbook?

    Can anyone recommend a textbook that goes beyond the usual techniques of integration taught in freshman/sophomore calculus and describes how to tackle more difficult integrals?

    In my reading, I'm encountering identities such as
    \int_{0}^{\infty} \frac{x dx}{e^{ax} + 1} = \frac{\pi^2}{12a^2}
    and
    \int_{0}^{2\pi} \frac{cos(\theta) d\theta}{A + B cos(\theta)} = \frac{2\pi}{B}(1 - \frac{A}{\sqrt{A^2 - B^2}})
    and while I am willing to believe that they are correct, I'd prefer to know where they come from.

    (I suspect the first can be solved using contour integration in the complex plane, and the second by a substitution such as u = tan(\theta / 2), but I haven't managed to work out the details of either and suspect that a good textbook would be helpful.)
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    Find the antiderivative of,
    \int\frac{x}{e^{ax}+1}dx,
    Write as,
    \int x(e^{ax}+1)^{-1}dx
    Now integrate by parts, u=x,v'=(e^{ax}+1)^{-1}
    The problem is finding what v is.

    Consider the integral,
    \int\frac{1}{e^{ax}+1}dx
    Write as,
    \frac{1}{a}\int\frac{ae^{ax}}{e^{ax}(e^{ax}+1)}dx
    Use substitution t=e^{ax} to get,
    \frac{1}{a}\int\frac{1}{t(t-1)}dt
    Decompose,
    \frac{1}{a}\int\frac{1}{t-1}-\frac{1}{t}dt
    Thus,
    \frac{1}{a}\left(\ln|t-1|-\ln|t|\right)+C
    Thus, v is,
    \frac{1}{a}\ln|1-e^{-ax}|+C
    -----
    Apply integration by parts,
    \frac{1}{a}x\ln|1-e^{-ax}|-\frac{1}{a}\int\ln|1-e^{-ax}|dx
    As, x\to\infty we have,
    \frac{1}{a}\int^{\infty}_0\ln|1-e^{-ax}|dx
    Which you can expess as, by simplifying
    \frac{1}{a^2}\int^{\infty}_0\ln|e^{ax}-1|dx
    But, I do not know if this helped.
    -------
    As for the second one I believe that Weirerstrauss substritution would be the trick. That is what you mentioned.
    Last edited by ThePerfectHacker; June 2nd 2007 at 07:02 PM.
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  3. #3
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    I follow what you've done, but I'm still not sure where the \frac{\pi^2}{12} comes from. I suspect that it's a contour integral because of the infinite series \frac{\pi^2}{12} = 1 - 1/4 + 1/9 - ... and the denominator of the integral gives such a nice sequence of poles along the imaginary axis. However, my complex analysis is much too rusty to actually make this work...

    However, I don't want to get bogged down in the details of these particular integrals... I'm more interested in learning exactly how to attack problems such as these in general. A good start would be a book that described "Weierstrass Substitution" and similar substitutions... I didn't realize that this substitution even had a name! (My undergrad calculus textbook ran through the usual arcsin and arctan substitutions and ended there... no hint of any other useful substitutions or how to apply complex analysis to the problem of finding real integrals.)

    Any suggestions?
    Last edited by TMFKAN64; March 10th 2006 at 04:21 PM.
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  4. #4
    TD!
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    Quote Originally Posted by TMFKAN64
    and the second by a substitution such as u = tan(\theta / 2)
    That's possible, although I'd apply a little trick to get rid off one cos, which will make the substitution easier. We rewrite

    \int {\frac{{\cos t}}{{a + b\cos t}}dt}  = \frac{1}{b}\int {\frac{{a + b\cos t - a}}{{a + b\cos t}}dt}

    So we simplify to

    \frac{1}{b}\int {dt}  - \frac{a}{b}\int {\frac{1}{{a + b\cos t}}dt}

    The first integral is easy of course, on the second you can apply that classic substitution. It will still be rather messy, because of the extra a in the denominator. You can work it towards an arctan (completing the square) but you surely won't get the final answer in the form which you have it.

    Because of the limits (0 to 2pi), this integral could be done using complex analysis as well, by using residue calculation. I'm assuming a > b > 0. The trick I just did to simplify the integral a bit is useful here too, and the first integral gives exactly that 2pi/b in your answer since

    \frac{1}{b}\int\limits_0^{2\pi } {dt}  = \frac{{2\pi }}{b}

    For the second (I'm leaving out the constant factor -a/b atm), I'll write

    I = \int\limits_0^{2\pi } {\frac{1}{{a + b\cos t}}dt}

    I substitute z = e^{it} so we integrate over the complex unity circle, letting t run from 0 to 2pi. In this substitution, we want cos(t) as a function of z which can be easily found using the definition of cos(t) for general complex values

    \cos t = \frac{{e^{it}  + e^{ - it} }}{2} = \frac{{z^2  + 1}}{{2z}}

    We also have that dt = dz/(iz), so our integral becomes

    I = \int\limits_0^{2\pi } {\frac{1}{{a + b\cos t}}dt}  = \oint\limits_{C^ +  } {\frac{1}{{a + b\frac{{z^2  + 1}}{{2z}}}}} \frac{{dz}}{{iz}}

    After some simplifying, we find

    \frac{1}{{bi}}\oint\limits_{C^ +  } {\frac{2}{{z^2  + 2\frac{a}{b}z + 1}}} dz

    Finding the zeroes of the denominator, we get

    z_{1,2} = \frac{{ - a \pm \sqrt {a^2  - b^2 } }}{b} = \frac{{ - a \pm b\sqrt {a^2 /b^2  - 1} }}{b}

    Let me define c = a/b > 0, so we get z_{1,2}  =  - c \pm \sqrt {c^2  - 1}
    Only the pole with the positive square root lies in our unity circle, so we only need the residue of our function there.

    Res \left( {\frac{2}{{z^2  + 2\frac{a}{b}z + 1}}, - c + \sqrt {c^2  - 1} } \right) = \frac{1}{{\sqrt {c^2  - 1} }}

    The, by the residue theorem

    I = 2\pi i\sum {Res}  = \frac{{2\pi i}}{{bi}}\frac{1}{{\sqrt {c^2  - 1} }} = \frac{{2\pi }}{b}\frac{1}{{\sqrt {a^2 /b^2  - 1} }}

    Combining with our initial 2pi/b and the factor -a/b I left out, we find

    \frac{{2\pi }}{b} - \frac{a}{b}\frac{{2\pi }}{b}\frac{1}{{\sqrt {a^2 /b^2  - 1} }} = \frac{{2\pi }}{b}\left( {1 - \frac{a}{{\sqrt {a^2  - b^2 } }}} \right)

    Which is, luckily, exactly what you had
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  5. #5
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    Quote Originally Posted by TMFKAN64
    I follow what you've done, but I'm still not sure where the \frac{\pi^2}{12} comes from. I suspect that it's a contour integral because of the infinite series \frac{\pi^2}{12} = 1 - 1/4 + 1/9 - ... and the denominator of the integral gives such a nice sequence of poles along the imaginary axis. However, my complex analysis is much too rusty to actually make this work...

    However, I don't want to get bogged down in the details of these particular integrals... I'm more interested in learning exactly how to attack problems such as these in general. A good start would be a book that described "Weierstrass Substitution" and similar substitutions... I didn't realize that this substitution even had a name! (My undergrad calculus textbook ran through the usual arcsin and arctan substitutions and ended there... no hint of any other useful substitutions or how to apply complex analysis to the problem of finding real integrals.)

    Any suggestions?
    During my free time I worked out this problem by hand. What you get is similar to,
    .....-\frac{1}{a}\int^{\infty}_0\ln(1+e^{ax})dx
    I now let a<0 thus, you can use infinite series,
    -\frac{1}{a}\int^{\infty}_0\left(e^{ax}-\frac{e^{2ax}}{2}+\frac{e^{3ax}}{3}-....\right)
    Integrate term-by-term,
    -\frac{1}{a^2}\left(e^{ax}-\frac{e^{2ax}}{2^2}+\frac{e^{3ax}}{3^2}+...\right)
    Evaluate the bounds, 0,\infty which gives,
    \frac{1}{a^2}\left(1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+...\right)
    Which gives,
    \frac{\pi^2}{12a^2}
    Also, the function in front of this integral which was obtained by integration by parts disappears when the limit at infinity is taken. Hope this helps

    It happens to by an extremely elegant problem. I do not know if this works with a>0.
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