Explicit formula for the series $ \sum_{k=1}^\infty \frac{x^k}{k!\cdot k} $

Question

I was wondering if there is an explicit formulation for the series $$ \sum_{k=1}^\infty \frac{x^k}{k!\cdot k} $$ It is evident that the converges for any $x \in \mathbb{R}$. Any ideas on a formula?

score 3 · Answer 1 · answered Jun 25 '13 at 10:20

3

Its derivative is $\sum_{k=1}^{\infty} \frac{x^{k-1}}{k!}$, which is $(e^x-1)/x$. So your function is the integral of my function, which you might or might not call 'closed form'.

answered Jun 25 '13 at 10:20

Empy2

50,853

Yes, but this is what we call the exponential integral! – Mhenni Benghorbal Jun 25 '13 at 10:29
When I was first-year undergrad, we were given so many examples of integrals that became arctans and logs that I thought 'closed form' mean exp,sin,cos,log, etc and no others. We didn't get Bessel, exponential integral etc until second year - but they appeared first as Taylor series so I thought you might be able to turn them into the functions I already knew. – Empy2 Jun 25 '13 at 13:05

Mhenni Benghorbal · Accepted Answer · 2013-06-25T10:39:16.480

2

You can have the closed form

$$\sum_{k=1}^{\infty}\frac{x^k}{k k!}= -\gamma-\ln(-x)-\Gamma(0, -x), $$

where $\Gamma(s,x)$ is the upper incomplete gamma function. Another possible form is

$$ \sum_{k=1}^\infty \frac{x^k}{k!\cdot k}=-\gamma-\ln \left( -x \right) -{\it Ei} \left( 1,-x \right), $$

where

$$ Ei(a, z) = \int_{1}^{\infty} \frac{e^{-tz}}{t^a}dt,\quad 0 < Re(z),$$

which is known as the exponential integral. The following relation between the exponential integral and the upper incomplete gamma function is useful

$$ Ei(a, z) = z^{a-1}\Gamma(1-a, z). $$

edited Jun 25 '13 at 10:39

answered Jun 25 '13 at 10:16

Mhenni Benghorbal

47,431

$\gamma$ is the Euler-Mascheroni constant? – Bunder Jun 25 '13 at 10:18
@Bunder: Yes, That's right. – Mhenni Benghorbal Jun 25 '13 at 10:19
And for positive $x$ ? – Bunder Jun 25 '13 at 10:24
You see that you have no problem for $Re(z)>0$. See the exponential integral. Then they are other means which you can use to extend the domain known as analytic continuation. – Mhenni Benghorbal Jun 25 '13 at 10:31
@Bunder: You are very welcome. – Mhenni Benghorbal Jun 25 '13 at 10:39

score 0 · Answer 3 · answered Jun 25 '13 at 10:43

Here is my attempt to unravel the mystery of the exponential integral. \begin{align*} e^x & = \sum_{k=0}^\infty \frac{x^k}{k!} \\ \frac 1xe^x & = \sum_{k=0}^\infty \frac{x^{k-1}}{k!} \\ \int_{1}^x \frac 1ye^y dy & = \sum_{k=1}^\infty \frac{x^k}{k! \cdot k} + \log x - \sum_{k=1}^\infty \frac{1}{k!k} \\ \int_{1}^x \frac 1ye^y dy - \log x + \sum_{k=1}^\infty \frac{1}{k!k} & = \sum_{k=1}^\infty \frac{x^k}{k!k} \end{align*} This derivation works for $x > 0$, but if $x < 0$, you only need to change the integral from $\int_1^x$ to $\int_{-1}^{x}$ and proceed from there.

Explicit formula for the series $ \sum_{k=1}^\infty \frac{x^k}{k!\cdot k} $

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