**(a)**

\begin{equation}

G(x,y,t) = \frac{1}{2 \sqrt{k \pi t}} e^{\frac{-(x-y)^2}{4kt}}\

\end{equation}

\begin{equation*}

u(x,t) = \int_{-\infty}^{\infty} G(x,y,t)g(y)dy \\

= \int_{-\infty}^{0} G(x,y,t)g(y)dy + \int_{0}^{\infty} G(x,y,t)g(y)dy \\

\end{equation*}

Let $-y = z$,

$$ = \int_{\infty}^{0} G(x,-z,t)g(-z)(-dz) + \int_{0}^{\infty} G(x,y,t)g(y)dy$$

We have Dirichlet boundary condition so g(z) is odd;$ g(-z) = -g(z).$

\begin{equation*}

= -\int_{\infty}^{0} G(x,-z,t)g(z)dz + \int_{0}^{\infty} G(x,y,t)g(y)dy\\

= -\int_{\infty}^{0} G(x,-y,t)g(y)dy + \int_{0}^{\infty} G(x,y,t)g(y)dy\\

= \int_{0}^{\infty}\big[G(x,y,t)g(y) - G(x,-y,t)g(y)\big]dy\\

= \int_{0}^{\infty} \frac{1}{2 \sqrt{k \pi t}} \left[ e^{\frac{-(x-y)^2}{4kt}} - e^{\frac{-(x+y)^2}{4kt}} \right] g(y)\,dy.\\

\end{equation*}

**(b)**

Neumann boundary conditions: $g(y)$ is even $\implies g(x) = g(-x)$

Making the same substitution as in **(a)**, we have:

\begin{equation*}

u(x,t) = \int_{0}^{\infty} \frac{1}{2 \sqrt{k \pi t}} \left[ e^{\frac{-(x-y)^2}{4kt}} + e^{\frac{-(x+y)^2}{4kt}} \right] g(y)\,dy.

\end{equation*}