Independency of solutions of the heat equation I'm facing a little setback with this exercise.
Given the heat equation
$$u_t=u_{xx}$$
and a generic solution $u(x,t)=t^{\alpha}\psi(k)$, where $k=\frac{x}{\sqrt t}$ and $\alpha$ is a constant, i need to show that 
$$\int_{-\infty}^{+\infty}u(x,t) dx = \int_{-\infty}^{+\infty}t^{\alpha}\psi(k) dx$$
Is independent of $t$ if and only if $\alpha=-\frac{1}{2}$.
I have adopted a straightforward approach: since $k=\frac{x}{\sqrt t}$, from this substitution immediately follows 
$$\int_{-\infty}^{+\infty}t^{\alpha}\psi(k)\sqrt t dk$$
From which the assertion immediately follows, since $t$ vanishes for that particular value of $\alpha$. Is there any error? Or it is correct? I hope that somebody can help!

By your deductions we see that
$$\begin{align}\int_{-\infty}^\infty u(x,t)\,dx&=\int_{-\infty}^\infty t^\alpha\psi(k)\,dx\\
&=\int_{-\infty}^\infty t^{\alpha+\frac{1}{2}}\psi(k)\,dk\\
&=t^{\alpha+\frac{1}{2}}\int_{-\infty}^\infty\psi(k)\,dk,
\end{align}$$
the latter is independent of $t$ if $\alpha=-\frac{1}{2}$. Assume $\alpha\ne -\frac{1}{2}$, then we also know that $u$ takes the general form (ignoring arbitrary constants in this case), $u(x,t)=t^{-\frac{1}{2}}e^{-\frac{x^2}{t}}$, and so $\psi(k)=t^{-\alpha-\frac{1}{2}}e^{-\frac{x^2}{t}}$, thus 
$$\int_{-\infty}^\infty\psi(k)\,dk=\int_{-\infty}^\infty t^{-\alpha-\frac{1}{2}}e^{-\frac{x^2}{t}}\,dk=t^{-\alpha-\frac{1}{2}}\int_{-\infty}^\infty e^{-k^2}\,dk\ne 0.$$
This tells thus that $\alpha\ne -\frac{1}{2}$ implies that the integral of $u$ does depend on $t$ (since the integral of $\psi$ is non zero), this gives us the contrapositive statement. Ultimately we have that the condition is if and only if.
