Applying the fourier transform to a PDE

[_Stew_]

I have a tutorial question for maths involving the heat equation and Fourier transform.

{\frac{∂u}{∂t}} = {\frac{∂^2u}{∂x^2}}

you are given the initial condition:

u(x,0) = 70e^{-{\frac{1}{2}}{x^2}}

the answer is:

u(x,t) = {\frac{70}{\sqrt{1+2t}}}{e^{-{\frac{x^2}{2+4t}}}}

In this course the definition of a Fourier transform is:

U(k) = F{u(x)}= {\frac{1}{\sqrt{2π}}}{\int_{-∞}^{∞}{{u(x)}{e^{-ikx}}}\,dx}

u(x) = F^{-1}{U(k)}= {\frac{1}{\sqrt{2π}}}{\int_{-∞}^{∞}{{U(k)}{e^{+ikx}}}\,dk}

MY SOLUTION:

So far I can use the transform with the pde to obtain
U(k,t) = A(k).e^{-k^{2}t}
U(k,0) = A(k)

From the initial condition and a transform table:
U(k,0) = 70.e^{-{\frac{1}{2}}k^2}
U(k,t) = 70.e^{-{\frac{1}{2}}k^2}.e^{-k^{2}t}

from here it looks like I would have to apply the inverse Fourier transform to get u(x,t). This is a difficult integration an I don't remember it being this complicated the first time. So basically:

Is there an easier way to do this question?
Any tips for the inverse Fourier integration below?

u(x,t) = {\frac{1}{\sqrt{2π}}}{\int_{-∞}^{∞}{{70.e^{-{\frac{1}{2}}k^2}.e^{-k^{2}t}}{e^{+ikx}}}\,dk}

 

[vela]

You have a Fourier transform of a Gaussian
$$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\alpha k^2} e^{ikx}\,dk$$ where $\alpha = t+\frac{1}{2}$, so you could just look up the answer in a table of Fourier transform pairs. If you want to derive the result, combine the exponentials and then complete the square.