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[itex] {\frac{∂u}{∂t}} = {\frac{∂^2u}{∂x^2}} [/itex]

you are given the initial condition:

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

the answer is:

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

In this course the definition of a fourier transform is:

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

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

MY SOLUTION:

So far I can use the transform with the pde to obtain

[itex] U(k,t) = A(k).e^{-k^{2}t}[/itex]

[itex] U(k,0) = A(k)[/itex]

From the initial condition and a transform table:

[itex] U(k,0) = 70.e^{-{\frac{1}{2}}k^2}[/itex]

[itex] U(k,t) = 70.e^{-{\frac{1}{2}}k^2}.e^{-k^{2}t}[/itex]

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?

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