I am trying to evaluate the following integral, which happens to be a laplace transform

$$ G(x,x';z)=\frac{1}{i\hslash}\int_0^{\infty}dt \exp\left(\frac{izt}{\hslash}\right)\sqrt{\frac{m}{2\pi i \hslash t}}\exp\left(-\frac{m(x-x')^2}{2i\hslash t}\right), $$ with $z=\frac{k^2\hslash^2}{2m}+i\epsilon$, where $\epsilon$ is positive, then taking the limit $\epsilon\rightarrow 0$.

I'm going to list my attempts and results. The answer to this integral should be $$ \frac{m}{ik\hslash^2}\exp{i k Abs(x)}. $$

Attempt 1:

2/(I \[HBar]) Integrate[E^(I ((z) u^2)/\[HBar]) Sqrt[m/(2\[Pi] I \[HBar])] E^(I (x^2 m)/(2\[HBar] u^2))/.z->((k \[HBar])^2/(2m)),{u,0,\[Infinity]},Assumptions->{k>0,\[HBar]>0,m>0,x\[Element]Reals,x!=0,z\[Element]Reals}]

Here I made the variable tranformation $t\rightarrow u^2$


If I set $z$ as real I get the correct answer.

If I set $z=\frac{k^2\hslash^2}{2m}+i\epsilon$

2/(I \[HBar]) Integrate[E^(I ((z) u^2)/\[HBar]) Sqrt[m/(2\[Pi] I \[HBar])] E^(I (x^2 m)/(2\[HBar] u^2))/.z->((k \[HBar])^2/(2m)+I \[Epsilon]),{u,0,\[Infinity]},Assumptions->{k>0,\[HBar]>0,m>0,x\[Element]Reals,x!= 0,\[Epsilon]>0}]

I get a cossine instead of the exponential. TrigToExp doesn't solve it, the answer is indeed incorrect.

Attempt 2(Laplace Transform then Full Simplify):

1/(I \[HBar]) LaplaceTransform[HeavisideTheta[t] Sqrt[m/(2\[Pi] I \[HBar] t)]Exp[(I x^2 m)/(2\[HBar] t)],t,-I(e+I \[Epsilon])/\[HBar]]

FullSimplify[%/.{e->(k^2 \[HBar]^2)/(2m),\[Epsilon]-> 0},{k>0,\[HBar]>0,m>0,x\[Element]Reals,x!= 0,z\[Element]Reals}]

Here I get the right result.

I wouldn't be worried if Mathematica couldn't evaluate the integrals above, but it seems to be doing so and giving the wrong output. What am I doing wrong?

  • $\begingroup$ Note that setting z reals goes againt the method, z should definetly be complex. But when I set z real is when I get the right answer, which is very weird. $\endgroup$ – Pedro Portugal Jun 9 '17 at 19:01

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