Jacobians--the absolute determinants of the change of variable function--appear formidable and can be complicated. Nevertheless, they are an essential and unavoidable part of the calculation of a multivariate change of variable. It would seem there's nothing for it but to write down a $k+1$ by $k+1$ matrix of derivatives and do the calculation.
There's a better way. It's shown at the end in the "Solution" section. Because the purpose of this post is to introduce statisticians to what may be a new method for many, much of it is devoted to explaining the machinery behind the solution. This is the algebra of differential forms. (Differential forms are the things that one integrates in multiple dimensions.) A detailed, worked example is included to help make this become more familiar.
Background
Over a century ago, mathematicians developed the theory of differential algebra to work with the "higher order derivatives" that occur in multi-dimensional geometry. The determinant is a special case of the basic objects manipulated by such algebras, which typically are alternating multilinear forms. The beauty of this lies in how simple the calculations can become.
Here's all you need to know.
A differential is an expression of the form "$dx_i$". It is the concatenation of "$d$" with any variable name.
A one-form is a linear combination of differentials, such as $dx_1+dx_2$ or even $x_2 dx_1 - \exp(x_2) dx_2$. That is, the coefficients are functions of the variables.
Forms can be "multiplied" using a wedge product, written $\wedge$. This product is anti-commutative (also called alternating): for any two one-forms $\omega$ and $\eta$,
$$\omega \wedge \eta = -\eta \wedge \omega.$$
This multiplication is linear and associative: in other words, it works in the familiar fashion. An immediate consequence is that $\omega \wedge \omega = -\omega \wedge \omega$, implying the square of any one-form is always zero. That makes multiplication extremely easy!
For the purposes of manipulating the integrands that appear in probability calculations, an expression like $dx_1 dx_2 \cdots dx_{k+1}$ can be understood as $|dx_1\wedge dx_2 \wedge \cdots \wedge dx_{k+1}|$.
When $y = g(x_1, \ldots, x_n)$ is a function, then its differential is given by differentiation:
$$dy = dg(x_1, \ldots, x_n) = \frac{\partial g}{\partial x_1}(x_1, \ldots, x_n) dx_1 + \cdots + \frac{\partial g}{\partial x_n}(x_1, \ldots, x_n) dx_n.$$
The connection with Jacobians is this: the Jacobian of a transformation $(y_1, \ldots, y_n) = F(x_1, \ldots, x_n) = (f_1(x_1, \ldots, x_n), \ldots, f_n(x_1, \ldots, x_n))$ is, up to sign, simply the coefficient of $dx_1\wedge \dots \wedge dx_n$ that appears in computing
$$dy_1 \wedge \cdots \wedge dy_n = df_1(x_1,\ldots, x_n)\wedge \cdots \wedge df_n(x_1, \ldots, x_n)$$
after expanding each of the $df_i$ as a linear combination of the $dx_j$ in rule (5).
Example
The simplicity of this definition of a Jacobian is appealing. Not yet convinced it's worthwhile? Consider the well-known problem of converting two-dimensional integrals from Cartesian coordinates $(x, y)$ to polar coordinates $(r,\theta)$, where $(x,y) = (r\cos(\theta), r\sin(\theta))$. The following is an utterly mechanical application of the preceding rules, where "$(*)$" is used to abbreviate expressions that will obviously disappear by virtue of rule (3), which implies $dr\wedge dr = d\theta\wedge d\theta = 0$.
$$\eqalign{
dx dy &= |dx\wedge dy| = |d(r\cos(\theta)) \wedge d(r\sin(\theta))| \\
&= |(\cos(\theta)dr - r\sin(\theta)d\theta) \wedge (\sin(\theta)dr + r\cos(\theta)d\theta| \\
&= |(*)dr\wedge dr + (*) d\theta\wedge d\theta - r\sin(\theta)d\theta\wedge \sin(\theta)dr + \cos(\theta)dr \wedge r\cos(\theta) d\theta| \\
&= |0 + 0 + r\sin^2(\theta) dr\wedge d\theta + r\cos^2(\theta) dr\wedge d\theta| \\
&= |r(\sin^2(\theta) + \cos^2(\theta)) dr\wedge d\theta)| \\
&= r\ dr d\theta
}.$$
The point of this is the ease with which such calculations can be performed, without messing about with matrices, determinants, or other such multi-indicial objects. You just multiply things out, remembering that wedges are anti-commutative. It's easier than what is taught in high school algebra.
Preliminaries
Let's see this differential algebra in action. In this problem, the PDF of the joint distribution of $(X_1, X_2, \ldots, X_{k+1})$ is the product of the individual PDFs (because the $X_i$ are assumed to be independent). In order to handle the change to the variables $Y_i$ we must be explicit about the differential elements that will be integrated. These form the term $dx_1 dx_2 \cdots dx_{k+1}$. Including the PDF gives the probability element
$$\eqalign{
f_\mathbf{X}(\mathbf{x},\mathbf{\alpha})dx_1 \cdots dx_{k+1} &\propto \left(x_1^{\alpha_1-1}\exp\left(-x_1\right)\right)\cdots \left(x_{k+1}^{\alpha_{k+1}-1}\exp\left(-x_{k+1}\right) \right)dx_1 \cdots dx_{k+1} \\
&= x_1^{\alpha_1-1}\cdots x_{k+1}^{\alpha_{k+1}-1}\exp\left(-\left(x_1+\cdots+x_{k+1}\right)\right)dx_1 \cdots dx_{k+1}.
}$$
(The normalizing constant has been ignored; it will be recovered at the end.)
Staring at the definitions of the $Y_i$ a few seconds ought to reveal the utility of introducing the new variable
$$Z = X_1 + X_2 + \cdots + X_{k+1},$$
giving the relationships
$$X_i = Y_i Z.$$
This suggests making the change of variables $x_i \to y_i z$ in the probability element. The intention is to retain the first $k$ variables $y_1, \ldots, y_k$ along with $z$ and then integrate out $z$. To do so, we have to re-express all the $dx_i$ in terms of the new variables. This is the heart of the problem. It's where the differential algebra takes place. To begin with,
$$dx_i = d(y_i z) = y_i dz + z dy_i.$$
Note that since $Y_1+Y_2+\cdots+Y_{k+1}=1$, then
$$0 = d(1) = d(y_1 + y_2 + \cdots + y_{k+1}) = dy_1 + dy_2 + \cdots + dy_{k+1}.$$
Consider the one-form
$$\omega = dx_1 + \cdots + dx_k = z(dy_1 + \cdots + dy_k) + (y_1+\cdots + y_k) dz.$$
It appears in the differential of the last variable:
$$\eqalign{
dx_{k+1} &= z dy_{k+1} + y_{k+1}dz \\
&= -z(dy_1 + \cdots + dy_k) + (1-y_1-\cdots -y_k)dz \\
&= dz - \omega.
}$$
The value of this lies in the observation that
$$dx_1 \wedge \cdots \wedge dx_k \wedge \omega = 0$$
because, when you expand this product, there is one term containing $dx_1 \wedge dx_1 = 0$ as a factor, another containing $dx_2 \wedge dx_2 = 0$, and so on: they all disappear. Consequently,
$$\eqalign{
dx_1 \wedge \cdots \wedge dx_k \wedge dx_{k+1} &= dx_1 \wedge \cdots \wedge dx_k \wedge dz - dx_1 \wedge \cdots \wedge dx_k \wedge \omega \\
&= dx_1 \wedge \cdots \wedge dx_k \wedge dz.
}$$
Whence (because all products $dz\wedge dz$ disappear),
$$\eqalign{
dx_1 \wedge \cdots \wedge dx_{k+1} &= (z dy_1 + y_1 dz) \wedge \cdots \wedge (z dy_k + y_k dz) \wedge dz \\
&= z^k dy_1 \wedge \cdots \wedge dy_k \wedge dz.
}$$
The Jacobian is simply $|z^k| = z^k$, the coefficient of the differential product on the right hand side.
Solution
The transformation $(x_1, \ldots, x_k, x_{k+1})\to (y_1, \ldots, y_k, z)$ is one-to-one: its inverse is given by $x_i = y_i z$ for $1\le i\le k$ and $x_{k+1} = z(1-y_1-\cdots-y_k)$. Therefore we don't have to fuss any more about the new probability element; it simply is
$$\eqalign{
&(z y_1)^{\alpha_1-1}\cdots (z y_k)^{\alpha_k-1}\left(z(1-y_1-\cdots-y_k)\right)^{\alpha_{k+1}-1}\exp\left(-z\right)|z^k dy_1 \wedge \cdots \wedge dy_k \wedge dz| \\
&= \left(z^{\alpha_1+\cdots+\alpha_{k+1}-1}\exp\left(-z\right) dz\right)\left( y_1^{\alpha_1-1} \cdots y_k^{\alpha_k-1}\left(1-y_1-\cdots-y_k\right)^{\alpha_{k+1}-1}dy_1 \cdots dy_k\right).
}$$
That is manifestly a product of a Gamma$(\alpha_1+\cdots+\alpha_{k+1})$ distribution (for $Z$) and a Dirichlet$(\mathbf\alpha)$ distribution (for $(Y_1,\ldots, Y_k)$). In fact, since the original normalizing constant must have been a product of $\Gamma(\alpha_i)$, we deduce immediately that the new normalizing constant must be divided by $\Gamma(\alpha_1+\cdots+\alpha_{k+1})$, enabling the PDF to be written
$$f_\mathbf{Y}(\mathbf{y},\mathbf{\alpha}) = \frac{\Gamma(\alpha_1+\cdots+\alpha_{k+1})}{\Gamma(\alpha_1)\cdots\Gamma(\alpha_{k+1})}\left( y_1^{\alpha_1-1} \cdots y_k^{\alpha_k-1}\left(1-y_1-\cdots-y_k\right)^{\alpha_{k+1}-1}\right).$$
I see two mistakes in your steps.
First of all, taking the derivative with respect to a single $\theta_m$ cancels out all the other terms where $i \ne m$ in the sum. That is,
$$
\frac{\partial}{\partial \theta_m} \sum_i (x_i + \alpha_i - 1) \log(\theta_i) = \frac{x_m + \alpha_m - 1}{\theta_m},
$$
as opposed to what you said. You now have $k$ equations to solve, one for each $m.$
Secondly, it's not enough to simply maximize the expression $\sum_i (x_i + \alpha_i - 1) \log(\theta_i).$ The solution is obviously $\theta_i \rightarrow \infty,$ because $\log$ monotonically increases. As you well know, we have the constraint that $\sum_i \theta_i = 1.$ You have to encode that constraint into your optimization via a Lagrange multiplier so that the correct solution isn't obviously infinity. That is, we must find the stationary point of
$$
L(\theta, \lambda) = \sum_{i=1}^k (x_i + \alpha_i - 1) \log(\theta_i) - \lambda \left[ \sum_{i=1}^k \theta_i - 1 \right].
$$
Taking the derivative of the above expression with respect to each $\theta_i$ and WRT the Lagrange multiplier $\lambda,$ and setting them to zero, you have to simultaneously solve,
$$
\frac{x_m + \alpha_m - 1}{\theta_m} = \lambda
$$
for each $m,$ and
$$
\sum_i \theta_i = 1.
$$
That's not very hard to do. Rearrange the first equation above to get an expression of $\theta_m$ in terms of $\lambda.$ Then, you can sensibly find the value of $\lambda$ that correctly imposes the constraint.
Best Answer
Just an initial remark, if you want computational speed you usually have to sacrifice accuracy. "More accuracy" = "More time" in general. Anyways here is a second order approximation, should improve on the "crude" approx you suggested in your comment above:
$$E\Bigg(\frac{X_{j}}{\sum_{i}X_{i}}\Bigg)\approx \frac{E[X_{j}]}{E[\sum_{i}X_{i}]} -\frac{cov[\sum_{i}X_{i},X_{j}]}{E[\sum_{i}X_{i}]^2} +\frac{E[X_{j}]}{E[\sum_{i}X_{i}]^3} Var[\sum_{i}X_{i}] $$ $$= \frac{\alpha_{j}}{\sum_{i} \frac{\beta_{j}}{\beta_{i}}\alpha_{i}}\times\Bigg[1 - \frac{1}{\Bigg(\sum_{i} \frac{\beta_{j}}{\beta_{i}}\alpha_{i}\Bigg)} + \frac{1}{\Bigg(\sum_{i} \frac{\alpha_{i}}{\beta_{i}}\Bigg)^2}\Bigg(\sum_{i} \frac{\alpha_{i}}{\beta_{i}^2}\Bigg)\Bigg] $$
EDIT An explanation for the above expansion was requested. The short answer is wikipedia. The long answer is given below.
write $f(x,y)=\frac{x}{y}$. Now we need all the "second order" derivatives of $f$. The first order derivatives will "cancel" because they will all involve multiples $X-E(X)$ and $Y-E(Y)$ which are both zero when taking expectations.
$$\frac{\partial^2 f}{\partial x^2}=0$$ $$\frac{\partial^2 f}{\partial x \partial y}=-\frac{1}{y^2}$$ $$\frac{\partial^2 f}{\partial y^2}=2\frac{x}{y^3}$$
And so the taylor series up to second order is given by:
$$\frac{x}{y} \approx \frac{\mu_x}{\mu_y}+\frac{1}{2}\Bigg(-\frac{1}{\mu_y^2}2(x-\mu_x)(y-\mu_y) + 2\frac{\mu_x}{\mu_y^3}(y-\mu_y)^2 \Bigg)$$
Taking expectations yields:
$$E\Big[\frac{x}{y}\Big] \approx \frac{\mu_x}{\mu_y}-\frac{1}{\mu_y^2}E\Big[(x-\mu_x)(y-\mu_y)\Big] + \frac{\mu_x}{\mu_y^3}E\Big[(y-\mu_y)^2\Big]$$
Which is the answer I gave. (although I initially forgot the minus sign in the second term)