---
This problem is a classic example of an integral in real analysis, and can be solved easily using complex analysis. We first extend the integrand, $\frac{(\ln x)^2}{x^2+1}$, to a holomorphic function $f(z)$ on the complex plane as follows:

$$f(z)=\frac{(\ln z)^2}{z^2+1}$$

Next, we transform the integration path $L$ into a large circle $\gamma$ centred at the origin with radius $R$. According to the theory of contour integration, as $R$ tends to infinity, the value of $\int_{\gamma} f(z)dz$ tends to 0. Therefore, we can break the integration path $L$ into three parts:

$$\int_{L} f(z)dz = \int_{0}^{R} \frac{(\ln x)^2}{x^2+1}dx + \int_{\gamma} f(z)dz - \int_{-R}^{0} \frac{(\ln x)^2}{x^2+1}dx$$

We are interested in computing $\int_{0}^{\infty} \frac{(\ln x)^2}{x^2+1}dx$, so we let $R$ tend to infinity. To evaluate $\int_{\gamma} f(z)dz$ and $\int_{-R}^{0} \frac{(\ln x)^2}{x^2+1}dx$, we first decompose $f(z)$ into:

$$f(z) = \frac{1}{2}\left(\frac{(\ln z + i\pi)^2}{z^2+1} - \frac{(\ln z - i\pi)^2}{z^2+1}\right)$$

Then, we use the residue theorem to evaluate $\frac{(\ln z + i\pi)^2}{z^2+1}$ and $\frac{(\ln z - i\pi)^2}{z^2+1}$ separately. Finally, we obtain the analytical expression for $\int_{0}^{\infty} \frac{(\ln x)^2}{x^2+1}dx$:

$$\int_{0}^{\infty} \frac{(\ln x)^2}{x^2+1}dx = \frac{\pi^3}{8}$$

This result is elegant and the method used is clever. Therefore, this problem serves as an excellent example for studying real analysis and complex analysis

$$\begin{equation}
\tag{1}
f(x) = \begin{cases}
\begin{aligned}
&\overline{f}, \quad && x \geq 0 \\
&1, \quad && x < 0 \\
&\underline{f}, \quad && x < 0
\end{aligned}
\end{cases}
\end{equation}$$
$f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$

The code first generates random data points from the two classes using the mvnrnd function. Then, it computes the discriminant function using the provided formula and evaluates it on a grid of points to plot the decision boundary. Finally, it uses the gscatter function to plot the distribution of the data points.

As for the provided code, the issue seems to be with the discriminant function. The formula for the discriminant function is incorrect. It should be:

$$g_i(x) = \ln(p_i) - \frac{1}{2}(x-\mu_i)^T\Sigma_i^{-1}(x-\mu_i) + \ln(c_{ij})$$

where $i$ is the class index, $j$ is the index of the other class, and $\Sigma_i$ is the covariance matrix for class $i$. The decision rule is then:

1 & g_1(x) > g_2(x) \\
2 & g_2(x) > g_1(x)
\end{cases}$$
Using this formula and the provided code, we can compute the discriminant function and plot the decision surface as follows:

根據給出的式子，我們需要求解 $\frac{\partial{\overline{{J}}}{m}}{\partial a{0}^{i}}$ 的值。