Помимо того, что в ответе многого не хватало, неясно, почему там не взяли второй интеграл — а он тоже Эйлера-Пуассона. Ответ получается вполне компактный.
Решение содержит много копипасты из решения №77, так что смотреть можно с середины.
Условия задачи:
\[
U=U\left(x,t\right),\quad-\infty < x < \infty,\quad t\geqslant0.
\]
\begin{equation}
U_{t}=a^{2}U_{xx}\label{eq:ur}
\end{equation}
\begin{equation}
\lim_{x\to\pm\infty}U=0\label{eq:usfur}
\end{equation}
\begin{equation}
\left.U\right|_{t=0}=U_{0}e^{-\frac{x^{2}}{l^{2}}}\label{eq:na4}
\end{equation}
Преобразования Фурье туда и оттуда:
\begin{equation}
\bar{f}\left(\lambda\right)=\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}f\left(\xi\right)e^{-i\lambda\xi}d\xi\qquad f\left(x\right)=\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}\bar{f}\left(\lambda\right)e^{i\lambda x}d\lambda,\quad\lim_{x\to\pm\infty}f\left(x\right)=0
\end{equation}
\begin{equation}
\bar{U}\left(\lambda,t\right)=\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}Ue^{-i\lambda x}dx
\end{equation}
Применим их к (\ref{eq:ur}):
\begin{equation}
U_{t}e^{-i\lambda x}=a^{2}U_{xx}e^{-i\lambda x}
\end{equation}
\begin{equation}
\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}U_{t}e^{-i\lambda x}dx=\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}a^{2}U_{xx}e^{-i\lambda x}dx
\end{equation}
\[
\frac{\partial}{\partial t}\left(\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}Ue^{-i\lambda x}dx\right)=\frac{a^{2}}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}\frac{\partial}{\partial x}U_{x}e^{-i\lambda x}dx=\left.\frac{a^{2}}{\sqrt{2\pi}}U_{x}e^{-i\lambda x}\right|_{-\infty}^{\infty}-\frac{a^{2}}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}U_{x}\frac{\partial}{\partial x}e^{-i\lambda x}dx=
\]
\begin{equation}
=\frac{i\lambda a^{2}}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}U_{x}e^{-i\lambda x}dx=\left.\frac{i\lambda a^{2}}{\sqrt{2\pi}}Ue^{-i\lambda x}\right|_{-\infty}^{\infty}-\frac{i\lambda a^{2}}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}U\frac{\partial}{\partial x}e^{-i\lambda x}dx=
\end{equation}
\[
-\lambda^{2}a^{2}\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}Ue^{-i\lambda x}dx
\]
\begin{equation}
\bar{U}_{t}=-\lambda^{2}a^{2}\bar{U}.
\end{equation}
Найдём Фурье-образ $\bar{U}$:
\begin{equation}
\ln\left|\bar{U}\right|=-\lambda^{2}a^{2}t+\tilde{\varphi}\left(\lambda\right),
\end{equation}
\begin{equation}
\bar{U}=\varphi\left(\lambda\right)e^{-\lambda^{2}a^{2}t}.
\end{equation}
Чтобы найти $\varphi\left(\lambda\right)$, преобразуем (\ref{eq:na4}):
\[
\left.\bar{U}\right|_{t=0}=\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}\left.U\right|_{t=0}e^{-i\lambda x}dx=
\]
\begin{equation}
=\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}U_{0}e^{-\frac{x^{2}}{l^{2}}}e^{-i\lambda x}dx=
\end{equation}
\[
=\frac{U_{0}}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}e^{-i\lambda x-\frac{x^{2}}{l^{2}}}dx.
\]
С другой стороны,
\begin{equation}
\left.\bar{U}\right|_{t=0}=\left.\varphi\left(\lambda\right)e^{-\lambda^{2}a^{2}t}\right|_{t=0}=\varphi\left(\lambda\right),
\end{equation}
\begin{equation}
\varphi\left(\lambda\right)=\frac{U_{0}}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}e^{-i\lambda\xi-\frac{\xi^{2}}{l^{2}}}d\xi.
\end{equation}
\begin{equation}
\bar{U}=\frac{U_{0}}{\sqrt{2\pi}}e^{-\lambda^{2}a^{2}t}\intop_{-\infty}^{\infty}e^{-i\lambda\xi-\frac{\xi^{2}}{l^{2}}}d\xi.
\end{equation}
Найдём саму функцию $U$:
\[
U=\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}\bar{U}e^{i\lambda x}d\lambda=
\]
\begin{equation}
=\frac{U_{0}}{2\pi}\intop_{-\infty}^{\infty}\intop_{-\infty}^{\infty}e^{-\lambda^{2}a^{2}t}e^{-i\lambda\xi-\frac{\xi^{2}}{l^{2}}}e^{i\lambda x}d\xi d\lambda=
\end{equation}
\[
=\frac{U_{0}}{2\pi}\intop_{-\infty}^{\infty}\intop_{-\infty}^{\infty}e^{-\lambda^{2}a^{2}t-i\lambda\xi-\frac{\xi^{2}}{l^{2}}+i\lambda x}d\lambda d\xi.
\]
Отдельно вычислим
\[
\intop_{-\infty}^{\infty}e^{-\lambda^{2}a^{2}t-i\lambda\xi-\frac{\xi^{2}}{l^{2}}+i\lambda x}d\lambda=\intop_{-\infty}^{\infty}e^{-\left[\left(\lambda a\sqrt{t}\right)^{2}-i\left(x-\xi\right)\lambda+\left(\frac{i\left(x-\xi\right)}{2a\sqrt{t}}\right)^{2}-\left(\frac{i\left(x-\xi\right)}{2a\sqrt{t}}\right)^{2}\right]-\frac{\xi^{2}}{l^{2}}}d\lambda=e^{-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}}\intop_{-\infty}^{\infty}e^{-\left[\lambda a\sqrt{t}-\frac{i\left(x-\xi\right)}{2a\sqrt{t}}\right]^{2}}d\lambda=
\]
$\lambda a\sqrt{t}-\frac{i\left(x-\xi\right)}{2a\sqrt{t}}=\gamma$,
$d\lambda=\frac{d\gamma}{a\sqrt{t}}$
\[
=e^{-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}}\intop_{-\infty}^{\infty}e^{-\gamma^{2}}\frac{d\gamma}{a\sqrt{t}}=\frac{\sqrt{\pi}}{a\sqrt{t}}e^{-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}}.
\]
Тогда
\[
U=\frac{U_{0}}{2\pi}\intop_{-\infty}^{\infty}\intop_{-\infty}^{\infty}e^{-\lambda^{2}a^{2}t-i\lambda\xi-\frac{\xi^{2}}{l^{2}}+i\lambda x}d\lambda d\xi=
\]
\[
=\frac{U_{0}}{2\pi}\intop_{-\infty}^{\infty}\frac{\sqrt{\pi}}{a\sqrt{t}}e^{-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}}d\xi=\frac{U_{0}}{2\sqrt{\pi}a}\frac{1}{\sqrt{t}}\intop_{-\infty}^{\infty}e^{-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}}d\xi.
\]
Примерно это было получено в методичке, но мы на этом не остановимся.
Преобразуем показатель
\[
-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}=-\frac{x^{2}-2x\xi+\xi^{2}}{4a^{2}t}-\frac{\xi^{2}}{l^{2}}=-\frac{x^{2}}{4a^{2}t}-\left[\left(\frac{1}{4a^{2}t}+\frac{1}{l^{2}}\right)\xi^{2}-2\frac{x}{4a^{2}t}\xi+\left(\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\right)^{2}-\left(\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\right)^{2}\right]=
\]
\[
=-\left[\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}\xi-\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\right]^{2}+\frac{\frac{x^{2}}{\left(4a^{2}t\right)^{2}}}{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}-\frac{x^{2}}{4a^{2}t}=
\]
\[
=-\left[\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}\xi-\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\right]^{2}+\frac{l^{2}}{l^{2}+4a^{2}t}\frac{x^{2}}{4a^{2}t}-\frac{x^{2}}{4a^{2}t}=
\]
\[
=-\left[\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}\xi-\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\right]^{2}-\frac{x^{2}}{l^{2}+4a^{2}t}
\]
Подставим в U и сведём к интегралу Эйлера-Пуассона
\[
U=\frac{U_{0}}{2\sqrt{\pi}a}\frac{1}{\sqrt{t}}\intop_{-\infty}^{\infty}e^{-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}}d\xi=\frac{U_{0}}{2\sqrt{\pi}a}\frac{1}{\sqrt{t}}e^{-\frac{x^{2}}{l^{2}+4a^{2}t}}\intop_{-\infty}^{\infty}e^{-\left[\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}\xi-\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\right]^{2}}d\xi=
\]
$\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}\xi-\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}=\gamma$,
$\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}d\xi=d\gamma$
\[
=\frac{U_{0}}{2\sqrt{\pi}a}\frac{1}{\sqrt{t}}e^{-\frac{x^{2}}{l^{2}+4a^{2}t}}\frac{1}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\intop_{-\infty}^{\infty}e^{-\gamma^{2}}d\gamma=\frac{U_{0}}{2a\sqrt{t}}\frac{e^{-\frac{x^{2}}{l^{2}+4a^{2}t}}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}.
\]
Ещё немного упростим:
\[
\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}=\sqrt{\frac{l^{2}+4a^{2}t}{4a^{2}tl^{2}}}=\frac{\sqrt{l^{2}+4a^{2}t}}{2a\sqrt{t}l},
\]
откуда
\[
U=\frac{U_{0}}{2a\sqrt{t}}\frac{e^{-\frac{x^{2}}{l^{2}+4a^{2}t}}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}=\frac{U_{0}}{2a\sqrt{t}}\frac{e^{-\frac{x^{2}}{l^{2}+4a^{2}t}}}{\frac{\sqrt{l^{2}+4a^{2}t}}{2a\sqrt{t}l}}=\frac{U_{0}l}{\sqrt{l^{2}+4a^{2}t}}e^{-\frac{x^{2}}{l^{2}+4a^{2}t}}.
\]
