# Integral to sum

1. May 12, 2008

### gop

1. The problem statement, all variables and given/known data

Show that

$$\int_{0}^{t}\frac{\lambda^{k}u^{k-1}e^{-\lambda u}}{(k-1)!}\ du$$

is

$$\sum_{j=k}^{\infty}\frac{(\lambda\cdot t)^{j}e^{-\lambda t}}{j!}$$

Hint: Use a taylor series to express e^(-lambda*u)

2. Relevant equations

3. The attempt at a solution

I used the taylor series to rewrite the e^ term and then I interchanges the integral sign with the summation and integrated the term which leaves me with

$$\sum_{j=0}^{\infty}\left[\frac{(\lambda\cdot t)^{j}}{j!}\cdot\frac{(\lambda\cdot t)^{k}}{(k+j)(k-1)!}\right]$$

However, I have no idea how to rewrite the left-erm to e^{-\lambda t} especially since we don't sum over k (I suspect it has something to do with shifting the index to j=k but I don't really know how that can be done without creating a second summation)

2. May 12, 2008

### e(ho0n3

The Taylor series of $e^{-\lambda u}$ will be an alternating series. Your result is not an alternating series.

But in any case, I suspect there may be a typo in the answer. You can shift the index by setting i = j + k and change the summand to work with i.

3. May 12, 2008

### gop

yeah you are right the (-1)^n is definitely missing.

Nevertheless, I tried your method of index shifting and I could even simplify the expression to some degree; however, I'm still not able to get a e^... in the answert. To get that I would need a taylor series or some expoential function I could rewrite in terms of e^... but how to do that is a complete mystery to me.

thx