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Math Help - differential equations

  1. #1
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    differential equations

    If a and b are positive constants, evaluate the following integrals:

    The integral from 0 to infinity, of (e^-ax sin bx)/x dx

    ta said answer was: pie/2 - tan^-1 (a/b), but book says tan^-1(b/a), so which? and how?

    The integral from 0 to infinity, of (e^-ax - e^-bx)/x dx

    ta said answer was ln(a/b), but book says log b/a, so which? and how?

    thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dixie View Post
    If a and b are positive constants, evaluate the following integrals:

    The integral from 0 to infinity, of (e^-ax sin bx)/x dx

    ta said answer was: pie/2 - tan^-1 (a/b), but book says tan^-1(b/a), so which? and how?

    The integral from 0 to infinity, of (e^-ax - e^-bx)/x dx

    ta said answer was ln(a/b), but book says log b/a, so which? and how?

    thanks!
    first, what does this have to do with differential equations? secondly, are you sure the /x should be there? example, \int e^{-ax} \sin bx ~dx can be worked out, but \int \frac {e^{-ax} \sin bx}x ~dx has no solution in elementary functions. it's an elliptical integral i believe

    also, be careful with what your TA says and make sure all the signs are right. for instance, -ln(a/b) = ln(b/a). so make sure you're not leaving off a minus sign or something
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  3. #3
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    Quote Originally Posted by dixie View Post
    If a and b are positive constants, evaluate the following integrals:

    The integral from 0 to infinity, of (e^-ax sin bx)/x dx
    First, \int e^{cx}\sin bx dx = \frac{e^{cx}}{c^2+b^2} \left( c\sin bx - b\cos x \right) + k for b,c\not = 0.

    Therefore, if c<0 then \int \limits_0^{\infty} e^{cx} \sin bx dx = \frac{e^{cx}}{c^2+b^2} \left( c\sin bx - b\cos x \right) \bigg|_0^{\infty} = \frac{b}{b^2+c^2}

    Also, \int_0^{\infty} e^{-xy} dy = \frac{1}{x} for x>0.

    Now, \int_0^{\infty} \frac{e^{-ax}\sin bx}{x} dx = \int \limits_0^{\infty} \int \limits_0^{\infty} e^{-ax}\sin (bx) e^{-xy} dy ~ dx

    Change integration order, \int \limits_0^{\infty} \int \limits_0^{\infty} e^{-x(a+y)}\sin (bx) dx ~ dy = \int \limits_0^{\infty} \frac{b}{b^2 + (a+y)^2}dy

    But, \int \limits_0^{\infty} \frac{b}{b^2 + (a+y)^2}dy = \int \limits_a^{\infty} \frac{b}{b^2 + t^2} dt = \tan^{-1}\frac{t}{b} \bigg|_a^{\infty} = \frac{\pi}{2} - \tan^{-1} \frac{a}{b}

    This is the answer the your TA got.
    Which is the same as \tan^{-1} \frac{b}{a} because \tan^{-1} \tfrac{1}{x} = \tfrac{\pi}{2} - \tan x,x>0
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    Quote Originally Posted by dixie View Post
    The integral from 0 to infinity, of (e^-ax - e^-bx)/x dx
    Again, \frac{1}{x} = \int_0^{\infty} e^{-yx} dy for x>0.

    Thus, \int_0^{\infty} \frac{e^{-ax} - e^{-bx}}{x} dx = \int \limits_0^{\infty} \int \limits_0^{\infty} \left( e^{-ax} - e^{-bx} \right) e^{-yx} dy ~ dx.

    Switch order, \int \limits_0^{\infty}\int \limits_0^{\infty} e^{-(a+y)x} - e^{-(b+y)x} dx ~ dy = \int \limits_0^{\infty} \frac{1}{a+y} - \frac{1}{b+y}dy = \log \left( \frac{a+y}{b+y} \right) \bigg|_0^{\infty}

    Thus, we get, \log 1 - \log \tfrac{a}{b} = - \log \tfrac{a}{b} = \log \tfrac{b}{a}
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  5. #5
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    It's actually quickier by considering \frac{{{e}^{-ax}}-{{e}^{-bx}}}{x}=\int_{a}^{b}{{{e}^{-xy}}\,dy}.

    And of course, the interchange of order of integration is justified by Tonelli's Theorem for non-negative functions.
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