The original problem is :
Let $\theta$ be a number such that $1/2<\theta<1$. Prove that
$$F(t,x)=\int_{-\infty}^{+\infty}\exp[i\tau t-(i\tau)^{1/2}x - (i\tau)^\theta] \,d\tau$$
defines a $C^\infty$ function in $\mathbb R^2$ which is a solution of the homogeneous heat equation
$$\frac{\partial F}{\partial t}=\frac{\partial^2 F}{\partial x ^2}\quad\text{ (in $\mathbb R^2$)}$$
I tried to simplify the integrand but got nothing. First of all, I wanted to show the integral converges, but I've no idea. Can anyone give me some ideas to start? Thanks!
Assuming that the powers $(i\tau)^\alpha$ use the principal branch, we have, for $\tau\neq 0$,
$$\begin{align} (i\tau)^{\alpha} &= \left((i\operatorname{sign} \tau) \lvert \tau\rvert\right)^\alpha\\ &= \left(e^{\operatorname{sign}\tau\cdot \pi i/2}\lvert \tau\rvert\right)^\alpha\\ &= \exp \left(\operatorname{sign}\tau \frac{\pi i \alpha}{2}\right) \lvert \tau\rvert^\alpha, \end{align}$$
and therefore
$$\operatorname{Re}\: (i\tau)^{\alpha} = \cos \left(\frac{\pi\alpha}{2}\right) \lvert \tau\rvert^\alpha.$$
Thus, since $\lvert e^z\rvert = e^{\operatorname{Re} z}$,
$$\left\lvert\exp \left[i\tau t - (i\tau)^{1/2} x - (i\tau)^\theta\right] \right\rvert = \exp \left[-x\sqrt{\lvert\tau\rvert}\cos\frac{\pi}{4} - \lvert\tau\rvert^\theta\cos \frac{\pi\theta}{2}\right].$$
Since $\frac{1}{2} < \theta < 1$, we have $\cos \frac{\pi\theta}{2} > 0$, and $\lvert\tau\rvert^\theta$ grows faster than $\lvert\tau\rvert^{1/2}$, hence for $\lvert x\rvert < K$, the integrand is dominated by
$$\exp \left(K \sqrt{\frac{\lvert\tau\rvert}{2}} - c\cdot \lvert\tau\rvert^\theta\right) \leqslant C\exp \left(-\tilde{c}\lvert\tau\rvert^\theta\right)\tag{$\ast$}$$
for some positive constants $C,\tilde{c}$ and $c = \cos \frac{\pi\theta}{2}$. The function on the right hand side of $(\ast)$ is integrable, and remains so when multiplied with arbitrary positive powers of $\lvert\tau\rvert$, so the dominated convergence theorem shows that $F$ is smooth, and the partial derivatives are obtained by differentiating under the integral. Differentiating under the integral also yields the differential equation, since $\frac{\partial}{\partial t}$ brings a factor of $i\tau$ down from the exponent, and each $\frac{\partial}{\partial x}$ a factor of $(i\tau)^{1/2}$.