# Integral of exponential function over an $n-1$ - simplex

I am trying to solve the following integral over the simplex (I'm not sure if there even is a closed form to be honest)

$$\int_{\Delta^{n-1}}\prod\limits_{i = 1}^n x_{i}^{a_i}e^{-b_ix_i}dx_i$$

Where $$a_i, b_i \in \mathbb{R}$$, $$a_i > -1, b_i \geq 0$$, $$x_i \in \left[0,1\right]$$, and $$\sum\limits_{i=1}^n x_i = 1$$.

My first plan of attack was to rewrite this integral as

$$\int\limits_0^1\int\limits_0^{1-x_{1}}\cdots\int\limits_0^{1-\sum\limits_{i=1}^{n-1} x_i}\prod\limits_{i = 1}^n x_{i}^{a_i}e^{-b_ix_i}dx_i$$ $$= \int\limits_0^1\int\limits_0^1\cdots\int\limits_0^1\delta\left(1-\sum\limits_{i=1}^{n-1} x_i\right)\prod\limits_{i = 1}^n x_{i}^{a_i}e^{-b_ix_i}dx_i$$ where I use the delta function to enforce the simplex. This allows me to use this definition of the delta function $$\delta\left(1-\sum\limits_{ i=1}^{n-1} x_i\right) = \int\limits_{-\infty}^{\infty}e^{-2\pi i\left(1-\sum\limits_{i=1}^{n-1} x_i\right)k}dk$$

And now I have a separable integral that looks like this $$\int\limits_{-\infty}^{\infty}\int\limits_0^1\int\limits_0^1\cdots\int\limits_0^1e^{-2\pi i\left(1-\sum\limits_{i=1}^{n-1} x_i\right)k}\prod\limits_{i = 1}^n x_{i}^{a_i}e^{-b_ix_i}dx_ idk$$ Now, I tried to start with the one-dimensional case, which should look like this $$\int\limits_{-\infty}^{\infty}\int\limits_0^1 x^{a}e^{-bx}e^{-2\pi i \left(1-x\right)k}dx dk$$

but doing the $$x$$ integral first gives me $$\int\limits_{-\infty}^{\infty}\int\limits_0^1 x^{a}e^{-bx}e^{-2\pi i \left(1-x\right)k}dx dk = \int\limits_{-\infty}^{\infty} \frac{\gamma\left(a-1, 1\right)}{\left(b+2\pi ik\right)^{a-1}}e^{-2\pi ik}dk \tag{1}$$ where $$\gamma\left(s,u\right)$$ is the lower incomplete gamma function. This integral is also equivalent to (assuming the unit step function $$H(0) = 1$$) $$\int\limits_0^1 \delta\left(1-x\right)x^{a}e^{-bx}dx = e^{-b} \tag{2}$$

The integral in $$(1)$$ is rather complicated and I don't know how to do it to get the result in $$(2)$$. Moreover, when I generalize to the $$n-$$dimensional case, I would have a product of these incomplete gamma functions in the integral which I don't really know how to work with.

Is this maybe the wrong approach to this integral? Does this integral over the simplex even have a closed form?

For context, this integral comes from putting a Dirichlet prior on $$x_i$$ for a product of Poisson distributions with mean $$b_ix_i$$

As suggested by Gérard Letac, we are looking at the Laplace transform of a Dirichlet distribution of parameter $$\alpha=(\alpha_1,\dotsc,\alpha_n)$$ where $$\alpha_i=a_i+1$$ (and the $$a_i$$'s are those in the question), computed at a point $$b=(b_1,\dotsc, b_n)$$.

The Fourier (hence the Laplace) transform $$\widehat{D_\alpha}(\mathbb{i}b)$$ of the Dirichlet distribution $$\widehat{D_\alpha}(s):= \int_{\Delta^{n-1}} \prod_{i=1}^n x_i^{\alpha_i-1} e^{\mathbb{i}\, s_i x_i} {\rm d}x_i, \qquad s=(s_1,\dotsc,s_n)\in\mathbb R^n,$$ is known in several equivalent formulations. It will be convenient to distinguish vectors from scalars, so vectors are from now on denoted by bold-face letters. $$\mathbb{i}$$ is the imaginary unit.

1. contracted (contour) integral representation$$\quad$$ Let $$B(a,b)$$ be the Euler beta function. Let $${}_nF_D[a,\mathbf{b},c,\mathbf{s}]:=\frac{1}{B(a,c-a)}\int_0^1 t^{a-1}(1-t)^{c-a-1} \prod_{i=1}^n (1-ts_i)^{-b_i}{\rm d}t, \qquad \Re(c)>\Re(a)>0,$$ be the integral representation of the $$n$$-variate hypergeometric Lauricella function of type $$D$$. The (first) confluent form of $${}_nF_D$$ is the second Kummer function $${}_n\Phi_2[\mathbf b;c;\mathbf s]:= \lim_{\epsilon\to 0^+} {}_kF_D[\epsilon^{-1};\mathbf b; c, \epsilon\, \mathbf s].$$

Then, $$\widehat{D_{\boldsymbol\alpha}}(\mathbf s)= {}_n\Phi_2[\boldsymbol\alpha;\alpha_1+\cdots+\alpha_n;\mathbb{i}\mathbf{s}].$$

1. hypergeometric series representation $$\quad$$ The second Kummer function also has a representation as hypergeometric series in terms of the Pochhammer symbol $$\langle \alpha \rangle_n:=\alpha(\alpha+1)(\alpha+2)\cdots (\alpha+n-1)$$, and we have

$$\widehat{D_{\boldsymbol\alpha}}(\mathbf s)= \sum_{\mathbf m\in \mathbb N_0^n} \frac{\langle\alpha_1\rangle_{m_1}\cdots \langle\alpha_n\rangle_{m_n}}{m_1!\cdots m_n!} \frac{\mathbb i^{m_1+\cdots+m_n} s_1^{m_1}\cdots s_n^{m_n}}{\langle \alpha_1+\cdots+\alpha_n\rangle_{m_1+\cdots+m_n}}.$$

1. representation as exponential generating function of cycle index polynomials $$\quad$$ Finally, the second Kummer function can also be expressed (and this is probably the most meaningful form) as the exponential generating function of (renormalized) cycle index polynomials of symmetric groups. Let $$Z_k$$ be the cycle index polynomial of the symmetric group acting on $$k$$ elements. Then, $$\widehat{D_{\boldsymbol\alpha}}(-\mathbb{i}\, t\, \mathbf s)={}_n\Phi_2[\boldsymbol\alpha; \alpha_1+\cdots+\alpha_n;t\mathbf{s}]= \sum_{k=0}^\infty \frac{t^k}{\langle\alpha_1+\cdots+\alpha_n\rangle_k} Z_k(\mathbf s^{\circ 1}\cdot\boldsymbol\alpha,\mathbf s^{\circ 2}\cdot\boldsymbol\alpha,\cdots, \mathbf s^{\circ k}\cdot\boldsymbol\alpha)$$ where $$\mathbf s^{\circ i} \cdot \boldsymbol\alpha:= \alpha_1s_1^i+\cdots +\alpha_n s_n^i$$ is the scalar product between $$\boldsymbol\alpha$$ and the $$i$$-th entry-by-entry (Hadamard) power of $$\mathbf s$$ and the explicit expression for $$Z_k$$ is $$Z_k[a_1,\cdots,a_k]= \sum_{j_1+2 j_2 + 3 j_3 + \cdots + k j_k = k} \frac{1}{\prod_{h=1}^k h^{j_h} j_h!} \prod_{h=1}^k a_h^{j_h}.$$

References and proofs for all the above statements can be found here (open access).

Remark Depending on the purpose of the computation, some of the above representations are more useful than others. The formulation in 3. is especially useful if you need $$n\gg 1$$. (Actually, this is seen already when $$n\geq 2$$.)

Remark Using the fact that the exponential generating function of the cycle index polynomials (this time non-renormalized) is an exponential (see here), it might be possible to show that no closed form exists for the Fourier transform of the Dirichlet distribution (i.e. $${}_k\Phi_2$$ is in general not resummable to a composition of elementary functions)

Why looking for the Laplace transform of a Dirichlet distribution, while even in dimension $$n=2$$ this is not elementary and are called the Kummer special functions? Laplace transforms are good for addition of independent random variables: and who cares for adding something concentrated on a tetraedron?

In the other hand, Dirichlet distributions have their moments quite easy to compute. Let me mention also the following integral, with the notations $$x_n=1-x_1-\cdots-x_{n-1}$$ and $$a=a_1+\cdots+a_n$$ $$\frac{\Gamma(a)}{\Gamma(a_1)\cdots \Gamma(a_n))}\int_{x_i>0, i=1,\ldots,n}\frac{x_1^{a_1-1}\cdots x_n^{a_n-1}}{(b_1x_1+\cdots+b_nx_n)^a}dx_1\ldots dx_{n-1}=\frac{1}{b_1^{a_1}\ldots b_n^{a_n}}.$$ Here we assume $$a_1,\ldots, a_n,b_1,\ldots,b_n>0$$

This post should have been in comments, rather than answers, but it was a bit too long. Hoping it will be useful anyway.

• The motivation for introducing the Fourier transform definition of the delta function was to decouple the $x_i$ integrals which would otherwise have rather complicated bounds imposed by the simplex. The delta function allows me to do the integral w.r.t. to $x_i$ from $0$ to $1$ directly and obtain the result I have in $(1)$. Commented Sep 9, 2023 at 15:41

The solution to the one-dimensional integral in is as follows:

\begin{align*} \int_0^1 x^a e^{-bx} e^{-2\pi i (1-x) k} dx &= \frac{\gamma(a + 1, b + 2\pi i k)}{b + 2\pi i k} \end{align*} where γ(a,x) is the incomplete gamma function.

This solution can be derived using a series expansion of the incomplete gamma function.

It is important to note that this solution is only valid if the incomplete gamma function is invertible.