# Difficult Laplace Transform Type of Integral

Good afternoon. I have the following integral that I need help integrating;

$$\mathrm{F}\left(x\right) = \int_{0}^{\infty}\mathrm{e}^{s\left(j - 1/x\right)}\, \left[\mathrm{T}_{N}\left(s\right)\right]^{k - j}\,\mathrm{d}s$$

where $$\mathrm{T}_{N}\left(s\right)$$ is the truncated exponential function $$\mathrm{T}_{N}\left(s\right) = \sum_{n = 0}^{N}\frac{s^{n}}{n!}$$

and $$j,k$$ are whole numbers.

I figured that tackling this using integration by parts is the best choice, but naturally this could take a while given that my choice of $$u_{1}=\mathrm{T}_{N}^{k - j}\left(s\right)$$ yields a $$\mathrm{d}u_{1}$$ of $$\mathrm{d}u_{1} = \left(k - j\right)\mathrm{T}_{N}^{k-j-1}\left(s\right) \mathrm{T}_{N - 1}\left(s\right)\,\mathrm{d}s$$ since $$\mathrm{d}\mathrm{T}_{N}\left(s\right)/\mathrm{d}s = \mathrm{T}_{N - 1}\left(s\right)$$. This does not seem to be simplifying the problem though as the next iteration would require me to then choose my $$u_{2}$$ to be $$u_{2} = \frac{1}{k - j}\,\frac{\mathrm{d}u_{1}}{\mathrm{d}s} = \mathrm{T}_{N}^{k - j - 1}\left(s\right) \mathrm{T}_{N - 1}\left(s\right)$$

Are there any other approaches to this without this drawn out IBP method or without considering expanding the truncated exponential ?.

• There was an answer here but it was deleted due to an objection I had... but I had an idea and if that person could put it back or "undelete it" it would help... I think I know how to resolve it... May 29 '19 at 14:55
• meta.stackexchange.com/questions/126348/… May 31 '19 at 1:56
• here it says that a moderator or 10k+ reputation account can view deleted answers. Maybe you can get one of them to help you put it up again? May 31 '19 at 1:57

I have only a partial answer: it pretty much just consists of repeated integration by parts, so maybe this isn't what you're looking for.

Write $$A(n_1, \dots, n_r) = \int_0^\infty e^{-as} \prod_i T_{n_i}(s) \,ds$$. Then integration by parts gives \begin{align*} A(n_1, \dots, n_r) &= -\frac{1}{a}\bigg[e^{-as}\prod_i T_{n_i}(s)\bigg]\bigg|_0^\infty + \frac{1}{a}\sum_{j : n_j > 0} \int_0^\infty e^{-as} T_{n_j - 1}(s)\prod_{i \neq j} T_{n_i}(s) \,ds \\ &= \frac{1}{a} + \frac{1}{a} \sum_{j : n_j > 0} A(n_1, \dots, n_j - 1, \dots, n_r) \end{align*} when all $$n_i \geq 0$$.

To make things a little easier, let's take $$T_n(s) = 0$$ when $$n < 0$$ (this is consistent with the identity $$T_n'(s) = T_{n-1}(s)$$), so we'll have $$A(n_1, \dots, n_r) = 0$$ when some $$n_i < 0$$, and so we can write $$A(n_1, \dots, n_r) = \frac{1}{a}\bigg(\mathbb{1}_{(n_i)} + \sum_i A(n_1, \dots, n_i - 1, \dots, n_r)\bigg)$$ for any integers $$n_1, \dots, n_r$$, where $$\mathbb{1}_{(n_i)} = 1$$ if all $$n_i \geq 0$$ and is $$0$$ otherwise.

Letting $$f(x_1, \dots, x_r) = \sum A(n_1, \dots, n_r) x_1^{n_1} \cdots x_r^{n_r}$$ be the generating function of $$A$$, this identity gives $$af = \frac{1}{(1-x_1) \cdots (1-x_r)} + (x_1 + \cdots + x_r)f$$ hence \begin{align*} f &= \frac{1}{(1 - x_1) \cdots (1 - x_r)(a - x_1 - \cdots - x_r)} \\ &= \frac{1}{a} \sum_{l_1, \dots, l_r, m_1, \dots, m_r} a^{-m_1 - \cdots - m_r}\binom{m_1 + \cdots + m_r}{m_1, \dots, m_r} x_1^{l_1 + m_1} \cdots x_r^{l_r + m_r} \\ &= \frac{1}{a} \sum_{n_1, \dots, n_r} \sum_{0 \leq m_i \leq n_i} a^{-m_1 - \cdots - m_r}\binom{m_1 + \cdots + m_r}{m_1, \dots, m_r} x_1^{n_1} \cdots x_r^{n_r} \\ \end{align*} and finally, $$A(n_1, \dots, n_r) = \frac{1}{a} \sum_{0 \leq m_i \leq n_i} a^{-m_1 - \cdots - m_r}\binom{m_1 + \cdots + m_r}{m_1, \dots, m_r}$$ (this could alternatively be found by repeatedly expanding the recurrence).

The case relevant to the question is the one where $$n_1 = \cdots = n_r = N$$, $$r = k-j$$, $$a = 1/x - j$$, but I'm not sure how to simplify this sum. I've been thinking about it in terms of lattice paths in $$r$$ dimensions: in particular the coefficient of $$a^{-t}$$ is the number of lattice paths of length $$t$$ where we have at most $$n_i$$ steps in the $$i$$-th direction for each $$i$$, but that doesn't seem to help simplify any further.

As noted by DinosaurEgg, we're only interested in $$A(N, \dots, N) = \frac{1}{a} \sum_{0 \leq m_i \leq N} a^{-m_1 - \cdots - m_r}\binom{m_1 + \cdots + m_r}{m_1, \dots, m_r}$$ but this sum still seems difficult to simplify to me, because of the $$m_i \leq N$$ constraints.

• I love what you did here, but I am giving the bounty to @ben, only because his work is closer to all that I have done with my calculations over the last couple of weeks. I definitely liked your approach though and will be referring to it as I continue. Thank you for your effort. May 31 '19 at 15:29

So long as $$j < 1/x$$ it is possible to write the function as a finite sum of terms (i.e., in closed form) by writing the power of the truncated exponential as a polynomial, and then turning the integral into a polynomial sum of gamma functions. If we apply the multinomial theorem we can write the power of the truncated exponential as a polynomial of degree $$N(k-j)$$ as follow:

\begin{aligned} \left[ \sum_{n=0}^N \frac{s^n}{n!} \right]^{k-j} &= \sum_{r_0 + \cdots + r_N = k-j} {k-j \choose \mathbf{r}} \prod_{n=0}^N \Big( \frac{s^n}{n!} \Big)^{r_n} \\[6pt] &= \sum_{r_0 + \cdots + r_N = k-j} \frac{(k-j)!}{\prod_{n=0}^N (r_n!) (n!)^{r_n}} \prod_{n=0}^N s^{n \cdot r_n} \\[6pt] &= \sum_{r_0 + \cdots + r_N = k-j} \frac{(k-j)!}{\prod_{n=0}^N (r_n!) (n!)^{r_n}} s^{\sum_{n=0}^N n \cdot r_n} \\[6pt] &= \sum_{i=0}^{N(k-j)} a_i s^i, \\[6pt] \end{aligned}

where the coefficients $$a_0,a_1,...,a_{N(k-j)}$$ depend on $$N$$ and $$k-j$$, and are obtained from this multinomial summation. We then have:

\begin{aligned} F(x) &= \int \limits_0^\infty e^{s (j-\frac{1}{x})} \left[ \sum_{n=0}^N \frac{s^n}{n!} \right]^{k-j} ds \\[6pt] &= \int \limits_0^\infty e^{s (j-\frac{1}{x})} \sum_{i=0}^{N(k-j)} a_i s^i \ ds \\[6pt] &= \sum_{i=0}^{N(k-j)} a_i \int \limits_0^\infty e^{s (j-\frac{1}{x})} s^i \ ds \\[6pt] \end{aligned}

Now, assume that $$j < 1/x$$ so that $$(j-\tfrac{1}{x}) <0$$. In this case we can use the change-of-variable $$m = s |j-\frac{1}{x}|$$ to get $$dm = -j ds$$ and we have:

\begin{aligned} F(x) &= \sum_{i=0}^{N(k-j)} a_i \int \limits_0^\infty e^{-s |j-\frac{1}{x}|} s^i \ ds \\[6pt] &= \sum_{i=0}^{N(k-j)} a_i \cdot \frac{1}{(-j) |j - \frac{1}{x}|^i} \int \limits_0^\infty e^{-s |j-\frac{1}{x}|} (s |j-\tfrac{1}{x}|)^i \ (-j) \ ds \\[6pt] &= \sum_{i=0}^{N(k-j)} a_i \cdot \frac{1}{(-j) |j - \frac{1}{x}|^i} \int \limits_0^\infty e^{-m} m^i \ dm \\[6pt] &= \sum_{i=0}^{N(k-j)} a_i \cdot \frac{i!}{(-j) |j - \frac{1}{x}|^i}. \\[6pt] \end{aligned}

This form of the function is a finite sum, where the only serious difficulty (for large $$N$$ or $$k-j$$) is to calculate the polynomial coefficients $$a_0,a_1,...,a_{N(k-j)}$$. Calculation of these coefficients may be cumbersome for large degree, but once they are calculated the function is then amenable to calculation.

• The answer I mentioned that was deleted had a similar expansion for the powered truncated exponential... I felt that it could be written as a sum of powers of $s$ better than that answerer did...I like where this is going... I suppose on first glance, those polynomial coefficients are hard to find a nice form for... Perhaps looking at small values of $N$ first and looking for pattern using computational methods could help here? May 31 '19 at 11:36
• Yes; I'm not sure of any simple recursion for the coefficients, but maybe some pattern could be found to assist this.
– Ben
May 31 '19 at 18:01

As mentioned in another answer deriving an exact expression is definitely hard, but we can bound this sum pretty easily from above:

$$A(N,..., N)<\frac{1}{a}\sum_{n=0}^{rN}\sum_{\sum m_i=n}a^{-n}\frac{n!}{m_1!...m_r!}=\frac{1}{a}\sum_{n=0}^{rN}\Big(\frac{r}{a}\Big)^n=\frac{\Big(\frac{r}{a}\Big)^{rN+1}-1}{r-a}=\frac{\Big(\frac{k-j}{1/x-j}\Big)^{rN+1}-1}{k-1/x}$$

since as user125932 mentions, the $$m_i$$'s are ranging up to $$N$$, and not $$rN$$ like the multinomial distribution would demand.

Also it is obvious that due to the fact that if we only sum up to $$n\leq N$$ we get an exact multinomial estimate and since all the summands are positive:

$$A(N,..., N)>\frac{1}{a}\sum_{n=0}^{N}\sum_{\sum m_i=n}a^{-n}\frac{n!}{m_1!...m_r!}=\frac{1}{a}\sum_{n=0}^{N}\Big(\frac{r}{a}\Big)^n=\frac{\Big(\frac{k-j}{1/x-j}\Big)^{N+1}-1}{k-1/x}$$

which by sandwiching reduces to the correct expression of the Laplace transform of the exponential in the limit $$N\rightarrow \infty$$:

$$\lim_{N\to\infty}A(N,..., N)=\frac{x}{1-kx}$$

To appreciate the complexity of this sum, let us attempt an evaluation for $$r=2$$. The sum is close to a binomial expansion but not exactly. We can break it up as follows (here $$\Delta=m+n$$): $$S_2=\sum_{m,n=0}^{N}{m+n\choose m}a^{-m-n}=\Big(\sum_{\Delta=0}^{N}\sum_{m=0}^{\Delta}+\sum_{\Delta=N+1}^{2N}\sum_{m=0}^{\Delta}-\sum_{\Delta=N+1}^{2N}\sum_{m=N+1}^{\Delta}\Big){\Delta\choose m}a^{-\Delta}$$

The first two terms in the sum can be summed from the binomial theorem and the third term is a remainder.

$$S_2=\frac{\Big(\frac{2}{a}\Big)^{2N+1}-1}{\frac{2}{a}-1}-R_2$$

By changing the order of summation we find for $$R_2$$

$$R_2=\sum_{m=0}^{N-1}\sum_{\Delta=0}^{N-1-m}{\Delta+N+m+1\choose\Delta}a^{-\Delta}<\sum_{m=0}^{N-1}\sum_{\Delta=0}^{\infty}{\Delta+N+m+1\choose\Delta}a^{-\Delta}=\sum_{m=0}^{N-1}(\frac{a}{a-1})^{m+N+1}=\frac{a^{N+1}}{(a-1)^N}((\frac{a}{a-1})^N-1)$$

(To be continued)

• This isn't quite the same though, right? My sum also constrains $m_i \leq N$, which is missing here -- I think this is what makes the sum difficult to work with. May 31 '19 at 3:22
• I agree with what you said. The approximation is good however for $N\to\infty$, so I'll try to do an analysis that explains why. Thanks for pointing it out! May 31 '19 at 4:45