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Math Help - integral challenge

  1. #1
    Eater of Worlds
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    integral challenge

    Here's one to give a go if you so desire. We always like a new integral to chew on, huh?.

    Evaluate:

    \int_{0}^{\infty}\frac{e^{ax}-e^{bx}}{x(e^{ax}+1)(e^{bx}+1)}dx

    a>b>0

    I know, I know, it's easy. Give it a go anyway.
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  2. #2
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    Quote Originally Posted by galactus View Post
    Here's one to give a go if you so desire. We always like a new integral to chew on, huh?.

    Evaluate:

    \int_{0}^{\infty}\frac{e^{ax}-e^{bx}}{x(e^{ax}+1)(e^{bx}+1)}dx

    a>b>0

    I know, I know, it's easy. Give it a go anyway.
    your integral = \int_0^{\infty} \int_b^a \frac{e^{xy}}{(e^{xy} + 1)^2} \ dy \ dx = \cdots = \frac{1}{2}\ln(a/b).
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  3. #3
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    You more than likely went about it easier than I.

    By letting f(x)=\frac{e^{x}}{e^{x}+1}+C

    I left f(ax)-f(bx)=\frac{e^{ax}-e^{bx}}{(e^{ax}+1)(e^{bx}+1)}dx

    Then setting I(u)=\int_{0}^{u}\frac{f(ax)}{x}dx-\int_{0}^{u}\frac{f(bx)}{x}dx

    Assuming they exist of course.

    Anyway, it took me longer than it did you, NCA. You are one sharp cat. I always like to peruse your posts for something useful and new to learn.
    You've forgotten more than I know.
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  4. #4
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    Quote Originally Posted by galactus View Post
    Here's one to give a go if you so desire. We always like a new integral to chew on, huh?.

    Evaluate:

    \int_{0}^{\infty}\frac{e^{ax}-e^{bx}}{x(e^{ax}+1)(e^{bx}+1)}dx

    a>b>0

    I know, I know, it's easy. Give it a go anyway.
    Another way is to define

     (1)\;\;\;I(m) = \int_0^\infty \frac{dx}{x \; \left( e^{mx} + 1 \right)}

    so your integral is given by  I(b)-I(a)

    If we differentiate (1) wrt m then

     I'(m) = \int_0^\infty \frac{- dx}{ \; \left( e^{mx} + 1 \right)^2}

    then integrate wrt x gives

     I'(m) = \frac{1}{ 2m} \;\;\; \Rightarrow\;\;\; I(m) = \frac{1}{2}\,\ln m + c

    Then,  I(b)-I(a) = \frac{1}{2} \ln \frac{b}{a} as given by NonCommAlg (well done btw)
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  5. #5
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    a shorter way is to notice that your integral is a Frullani's Integral if you let f(x)=\frac{1}{e^x + 1}.
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  6. #6
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    Cool Observation. I must admit, I should have saw that.
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  7. #7
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    Besides the above you could see right away that we can discompose our integral as follows

    \frac{x-y}{(x+1)(y+1)}=\frac{1}{y+1}-\frac{1}{x+1}\implies \frac{e^{ax}-e^{bx}}{(e^{ax}+1)(e^{bx}+1)}=\frac{1}{e^{bx}+1}-\frac{1}{e^{ax}+1}

    So our integral becomes

    \int_0^{\infty}\left\{\frac{1}{x(e^{bx}+1)}-\frac{1}{x(e^{ax}+1)}\right\}dx

    Now using Frullani's Theorem which states that assuming f'\in\mathcal{C} that \int_0^{\infty}\frac{f(bx)-f(ax)}{x}=\left(f(0)-f(\infty)\right)\ln\left(\frac{a}{b}\right)

    So if we let f(x)=\frac{1}{e^x+1}

    We get \int_0^{\infty}\left\{\frac{1}{x(e^{bx}+1)}-\frac{1}{x(e^{ax}+1)}\right\}dx=\left(\frac{1}{e^0  +1}-\frac{1}{e^{\infty}+1}\right)\ln\left(\frac{a}{b}\  right)=\frac{1}{2}\ln\left(\frac{a}{b}\right)


    EDIT: NonCommAlg...you ruin all my fun.
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  8. #8
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    I once saw the following on a Putnam exam

     \int_0^{\infty} \frac{\tan^{-1} a x - \tan^{-1} b x }{x}\; dx

    What a wonderful generalization in the link you provide - thanks for sharing!
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by danny arrigo View Post
    I once saw the following on a Putnam exam

     \int_0^{\infty} \frac{\tan^{-1} a x - \tan^{-1} b x }{x}\; dx

    What a wonderful generalization in the link you provide - thanks for sharing!
    This was in the Putnam . That is not on the same level as the Putnam integrals I have seen.
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  10. #10
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    Quote Originally Posted by Mathstud28 View Post
    This was in the Putnam . That is not on the same level as the Putnam integrals I have seen.
    the point is that a (good) first-year undergrad student is usually much better at intergration than a second-year student and things get worse and worse as you get a higher education in math.

    that's why most graduate students totally suck at it!
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    the point is that a (good) first-year undergrad student is usually better at intergration than a second-year student and things get worse and worse as you get a higher education in math.

    that's why most graduate students totally suck at it!
    Haha thats true...but still I found some integrals I have seen from past Putnams:

    Without CA calculate

    \int_{-\infty}^{\infty}\frac{x^3\sin(x)}{(1+x^2)^2}

    \int_0^{\infty}\frac{\cos(kx)}{1+x+x^2}dx~~k\geqsl  ant 0


    They are a lot of work....especially the first one...if I remember correctly when I did it I used Leibniz's Rule THREE times...
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    Quote Originally Posted by NonCommAlg View Post
    the point is that a (good) first-year undergrad student is usually much better at intergration than a second-year student and things get worse and worse as you get a higher education in math.

    that's why most graduate students totally suck at it!
    Well you know how the saying goes - a mile wide and a foot deep and a mile deep and a foot wide. So, after many years of study, I know a tremendous amount about absolutley nothing.
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  13. #13
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    Quote Originally Posted by Mathstud28 View Post
    Haha thats true...but still I found some integrals I have seen from past Putnams:

    Without CA calculate

    \int_{-\infty}^{\infty}\frac{x^3\sin(x)}{(1+x^2)^2}

    They are a lot of work....especially the first one...if I remember correctly when I did it I used Leibniz's Rule THREE times...
    let me guess! the answer is \frac{\pi}{2e} ?
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    let me guess! the answer is \frac{\pi}{2e} ?
    You mean \frac{\pi}{2e} and yes....but I sense some Complex Analysis in the mix here

    EDIT: You corrected yourself.
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  15. #15
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    Quote Originally Posted by NonCommAlg View Post
    let me guess! the answer is \frac{\pi}{2e} ?
    Simon Fraser University - I once went to high school with a Dave Muraki - know him?
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