# Double summation with improper integral

So my friend sent me this really interesting problem. It goes:

Evaluate the following expression:

$$\sum_{a=2}^\infty \sum_{b=1}^\infty \int_{0}^\infty \frac{x^{b}}{e^{ax} \ b!} \ dx .$$

Here is my approach:

First evaluate the integral:

$$\frac{1}{b!} \int_0^\infty \frac{x^b}{e^{ax}}\ dx.$$

This can be done using integration by parts and we get:

$$\frac{1}{b!} \frac{b}{a} \int_0^\infty \frac{x^{b-1}}{e^{ax}}\ dx.$$

We can do this $$b$$ times until we get:

$$\frac{1}{b!} \frac{(b)(b-a).....(b-b+1)}{a^b} \int_0^\infty \frac{x^{b-b}}{e^{ax}}\ dx.$$

and hence we end up with:

$$\frac{1}{b!} \frac{b!}{a^b}\qquad\left(\frac{-1 \ e^{-ax}}{a}\Big|_0^\infty\right) = \frac{1}{a^{b+1}}.$$

Now we can apply the sum of GP to infinity formula and we get:

$$\sum_{a=2}^\infty \sum_{b=1}^\infty \frac{1}{a^{b+1}} = \sum_{a=2}^\infty \frac{\frac{1}{a^{2}}}{1-\frac{1}{a}}.$$

This is a telescoping series and we end up with $$\frac{1}{a-1} = \frac{1}{2-1} = 1.$$

Do you guys have any other ways of solving this problem? Please do share it here.

• I think this is the simplest way have you done! Commented Oct 1, 2018 at 9:39
• One approach: interchange summations and integration (allowed by Tonelli's theorem) and perform the sum's ($e^x$ and geometrical series) to end up with a simple integral. Commented Oct 1, 2018 at 9:58

since $$\frac{x^b}{e^{ax} b!}$$ is non-negative, Tonelli's theorem for iterated integrals/sums allows us to interchange integrals and sums without worry. Then: \begin{align} &\sum_{a=2}^\infty \sum_{b=1}^\infty \int_{0}^\infty \frac{x^{b}}{e^{ax} \ b!} \ dx \\ &=\int_{0}^\infty \sum_{a=2}^\infty e^{-ax} \sum_{b=1}^\infty \frac{x^{b}}{ \ b!} \ dx \\ &= \int_{0}^\infty \underbrace{\left(\sum_{a=2}^\infty (e^{-x})^a\right)}_{\text{geometric series}} \overbrace{\left(\sum_{b=0}^\infty \frac{x^{b}}{ \ b!}-1\right)}^{\text{series definition of e^x}} \ dx \\ &= \int_{0}^\infty \frac{1}{e^x(e^x-1)}(e^{x}-1)dx \\ &= \int_0^\infty e^{-x} dx \\&= 1.\end{align}

The integral is of the Gamma type,

$$\int_{0}^\infty \frac{x^{b}}{e^{ax}} \ dx=\frac1{a^{b+1}}\int_{0}^\infty t^be^{-t}\ dx =\frac{b!}{a^{b+1}}.$$

Then

$$\sum_{a=2}^\infty \sum_{b=1}^\infty\frac1{a^{b+1}}=\sum_{a=2}^\infty \frac1{a^2\left(1-\dfrac1a\right)}=\sum_{a=2}^\infty \frac1{a(a-1)}$$ is indeed a telescoping sum, giving

$$\frac1{2-1}.$$

• Ah, it's interesting how you were able to link the integral to the gamma function.
– user271938
Commented Oct 1, 2018 at 12:23
• @Tusky: this is actually how the Gamma function is defined.
– user65203
Commented Oct 1, 2018 at 12:27

For the first part I often use the Laplace transform: $$\frac{1}{b!} \int_0^\infty \frac{x^b}{e^{ax}}\ dx = \frac{1}{b!} \int_0^\infty x^be^{-ax}\ dx = \frac{1}{b!} {\cal L}(x^b)\Big|_{s=a} = \frac{1}{b!} \frac{b!}{a^{b+1}} = \frac{1}{a^{b+1}}$$ this make it easier.

• And how do you evaluate the Laplace transform? I don't see how it's easier unless you 'cheat' and look up the result in a table. Commented Oct 1, 2018 at 9:52
• Laplace transform is a tool. We use it's formulas whatever we want! Commented Oct 1, 2018 at 10:01
• @Winther: there's actually a lot of ways to evaluate that. Integration by parts, differentiation under the integral sign (probably the easiest one), induction etc.
– edmz
Commented Oct 1, 2018 at 18:26

Also note: $$S=\sum_{a=2}^\infty\sum_{b=1}^\infty\int_0^\infty\frac{x^b}{e^{ax}b!}dx=\sum_{a=2}^\infty\sum_{b=1}^\infty\frac{1}{b!}\int_0^\infty x^be^{-ax}dx$$ now let: $$u=ax$$ $$dx=\frac{du}{a}$$ so: $$S=\sum_{a=2}^\infty\sum_{b=1}^\infty\frac{1}{a*b!}\int_0^\infty \left(\frac{u}{a}\right)^be^{-u}du=\sum_{a=2}^\infty\sum_{b=1}^\infty\frac{a^{-(b+1)}}{b!}\int_0^\infty u^be^{-u}du$$ and we know that: $$(n-1)!=\Gamma(n)=\int_0^\infty e^{-t}t^{n-1}dt$$ so our summation now simplifies to: $$S=\sum_{a=2}^\infty\sum_{b=1}^\infty\frac{a^{-(b+1)}}{b!}b!=\sum_{a=2}^\infty\sum_{b=1}^\infty\frac{1}{a^{b+1}}=\sum_{a=2}^\infty\sum_{c=2}^\infty\frac{1}{a^{c}}=\sum_{a=2}^\infty\left(\sum_{c=1}^\infty\frac{1}{a^c}-\frac{1}{a}\right)=\sum_{a=2}^\infty\sum_{c=1}^\infty\frac{1}{a^c}-\sum_{a=2}^\infty\frac{1}{a}$$ I know this is correct up to the second summation on the final line but after this I am not sure.

• You definitely don't want to pull out $\sum_2^\infty \frac1{a} = \infty$; note that the converse of $$\sum a_n < \infty , \sum b_n < \infty \implies \sum (a_n + b_n) < \infty$$ does not hold. Luckily you don't need to split the sum in $a$, since $$\sum_{c=1}^\infty \frac1{a^c} = \frac{1}{a-1}$$ so you have a telescoping sum, giving the result. Commented Oct 2, 2018 at 20:34