This problem arises from trying to solve, by Fourier transform, the Cauchy problem
$$\begin{cases}
u_{tt}-u_{xxxx}=0 &x\in\mathbb{R},\, t\geq 0\\
\begin{cases}
u(0,x)=f(x)\\
u_t(0,x)=0
\end{cases}
\end{cases}$$
That is fourth-order wave PDE.
Doing formal integrals, I arrive at this expression after completing squares and using definitions of error functions:
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\cosh(k^2t)e^{ikx}~dk=\frac{1}{4\sqrt{2t}}e^{-\frac{x^2}{4t}}\left[\text{erf}\left(\sqrt{t}\left(k-\frac{ix}{2t}\right)\right)+e^{\frac{x^2}{2t}}\text{erfi}\left(\sqrt{t}\left(k+\frac{ix}{2t}\right)\right)\right]_{-\infty}^{\infty}=\frac{1}{4\sqrt{2t}}e^{-\frac{x^2}{4t}}\biggl
(2+e^{\frac{x^2}{2t}}\left[\text{erfi}\left(\sqrt{t}\left(k+\frac{ix}{2t}\right)\right)\right]_{-\infty}^{\infty}\biggr)$$
where I have evaluated $\text{erf}$ in the last line. My problem is that I know that $\text{erfi}$ does not have a limit when $k\to\infty$ since it is non-bounded over real line. However doing this inverse Fourier transform using corresponding Mathematica command gives me:
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\cosh(k^2t)e^{ikx}~dk=\frac{1}{4\sqrt{2t}}e^{-\frac{x^2}{4t}}\left(2+2ie^{\frac{x^2}{2t}}\right)$$
where I have arranged the terms in order to get the same form as the result I have described before.
What am I not understanding?
Best Answer
It must be an error, because this Fourier integral doesn't even converge as tempered distributions, let alone as functions.
In general, anything of exponential growth or more doesn't have a Fourier transform in any sense.