Solving integral using feynman trick

Question

For $x,t > 0 \ $ I want to prove $$\int_0^\infty \frac{\sin(x n)}{x \ n} e^{-t n^2} dn = \text{erf} \left(\frac{x}{2 \sqrt{t}}\right) $$ using the feynman trick. My problem is I don't know which function to use it on. For example if I want to find $f(u)$ such that $$ \frac{\partial}{\partial u} f(u) = \frac{e^{- u t n^2}}{n} $$ I would have to take a principal value, which can't be helpful. Does anyone have a tip?

Answer 1

Define a function $g$ by $$g(n,x,t)=\frac{\sin(x n)}{xn} e^{-t n^2}$$ for $n,x,t > 0$. Now, $$\frac{\partial g}{\partial t}(n,x,t)=-n\frac{\sin(x n)}{x} e^{-t n^2}$$ Therefore $$\int_0^\infty \frac{\partial g}{\partial t}(n,x,t) dn = -\frac1{2x}\int_0^\infty \sin(nx)e^{-tn^2} 2ndn \\=-\frac1{2x}\int_0^\infty \sin(\sqrt{n}x)e^{-tn} dn$$ By the Laplace transform of $\sin(\sqrt{n}x)$, we have $$\frac1{x}\mathcal{L}\{\sin(\sqrt{n}x)\}(t)=\frac1{x}\int_0^\infty \sin(\sqrt{n}x)e^{-tn} dn=\frac{\sqrt{\pi}e^{-x^2/4t}}{2t^{\frac32}}$$ Now since $$-\frac{\partial }{\partial t}\int_0^\infty \frac{\sin(x n)}{xn} e^{-t n^2} dn =\frac{\sqrt{\pi}e^{-x^2/4t}}{4t^{\frac32}} $$ you can get the result finally beacuse $$\frac{\partial }{\partial t}\text{erf} \left(\frac{x}{2 \sqrt{t}}\right)=-\frac{x e^{-x^2/4t}}{2\sqrt{\pi}t^{\frac32}}$$ and $$\lim_{t\to\infty}\text{erf} \left(\frac{x}{2 \sqrt{t}}\right)=\text{erf}(0)=0\text{ for all }x>0$$