How can I deduce that
1√4πt∫∞−∞sin(y)e−(x−y)24tdy=e−tsin(x)
without actually evaluating the definite integral?
Answer
I will assume the following result:
The solution to the Heat Equation
dfdt=∇2f
with initial condition f(x,0)=g(x) can be written
f(x,t)=1√4πt∫∞−∞e−(x−y)24tg(y)dy
Now by inserting
f(x,t)=e−tsin(x)
into the Heat Equation we find that it does satisfy it with the initial condition f(x,0)=sin(x). From the result above it therefore follows that
e−tsin(x)=1√4πt∫∞−∞e−(x−y)24tsin(y)dy
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