Exponential Integration

**qwerty10** · November 22nd 2010, 05:15 AM

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

Tried a few different substitutions and can't seem to get any to work

**TheCoffeeMachine** · November 22nd 2010, 05:29 AM

$\displaystyle \int\frac{e^x-1}{x}\;{dx} = \int\left(\frac{e^x}{x}-\frac{1}{x}\right)\;{dx} = \int\frac{e^x}{x}}\;{dx}}-\int\frac{1}{x}\;{dx}} = Ei(x)-\log{x}+k$ .

Probably a pattern to remember (easy enough):

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

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

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

$\displaystyle \int\frac{e^x}{x}\;{dx} = Ei(x)+k.$

**qwerty10** · November 22nd 2010, 05:37 AM

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?

**TheCoffeeMachine** · November 22nd 2010, 07:34 AM

Originally Posted by qwerty10

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:

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

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

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

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