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Math Help - Exponential Integration

  1. #1
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    Exponential Integration

    I am having problems with integrating (e^x-1)/x.

    Tried a few different substitutions and can't seem to get any to work
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  2. #2
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    \displaystyle \int\frac{e^x-1}{x}\;{dx} = \int\left(\frac{e^x}{x}-\frac{1}{x}\right)\;{dx} = \int\frac{e^x}{x}}\;{dx}}-\int\frac{1}{x}\;{dx}} = Ei(x)-\log{x}+k.

    Probably a pattern to remember (easy enough):

    \displaystyle \int\frac{\sin{x}}{x}}\;{dx} = Si(x)+k.

    \displaystyle \int\frac{\cos{x}}{x}}\;{dx} = Ci(x)+k.

    \displaystyle \int\frac{\log{x}}{x}\;{dx} = Li(x)+k.

    \displaystyle \int\frac{e^x}{x}\;{dx} = Ei(x)+k.
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  3. #3
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    Thank you

    I now have to integrate between 0 to 1, for Ei(x) it would it not work? i.e I would get something divided by 0?
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  4. #4
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    Quote Originally Posted by qwerty10 View Post
    Thank you

    I now have to integrate between 0 to 1, for Ei(x) it would it not work? i.e I would get something divided by 0?
    Well, then you will need to find:

    \displaystyle L =  \lim_{x \to 1}\bigg\{\mathrm{Ei}(x)-\log(x)\bigg\}-\lim_{x \to 0}\bigg\{\mathrm{Ei}(x)-\log(x)\bigg\}.

    Then using the series \mathrm{Ei}(x) = \gamma+\log{x}+\sum_{k=1}^{\infty} \frac{x^k}{k\; k!} gives:

    \displaystyle L =  \lim_{x \to 1}\bigg\{\gamma+\log{x}+\sum_{k=1}^{\infty} \frac{x^k}{k\; k!}-\log(x)\bigg\}-\lim_{x \to 0}\bigg\{\gamma+\log{x}+\sum_{k=1}^{\infty} \frac{x^k}{k\; k!}-\log(x)\bigg\} .

    \displaystyle  =  \lim_{x \to 1}\bigg\{\gamma+\sum_{k=1}^{\infty} \frac{x^k}{k\; k!}\bigg\}-\lim_{x \to 0}\bigg\{\gamma+\sum_{k=1}^{\infty} \frac{x^k}{k\; k!}\bigg\}  = \gamma +\lim_{x \to 1}\sum_{k=1}^{\infty} \frac{x^k}{k\; k!}-\gamma-\lim_{x \to 0}\sum_{k=1}^{\infty} \frac{x^k}{k\; k!} .

    \displaystyle  = \lim_{x \to 1}\sum_{k=1}^{\infty} \frac{x^k}{k\; k!}-\lim_{x \to 0}\sum_{k=1}^{\infty} \frac{x^k}{k\; k!} = \lim_{x \to 1}\sum_{k=1}^{\infty} \frac{x^k}{k\; k!} = \lim_{x \to 1}\bigg(\mathrm{Ei(x)}-\gamma+\log{x}\bigg)   = \mathrm{Ei}(1)-\gamma.
    Last edited by TheCoffeeMachine; November 22nd 2010 at 07:52 AM.
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