# Solving integral using feynman trick

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?

• Mabye I should try to solve $\frac{sin(x)}{x}$ first? Also I tried to substitute $$\eta := n \ 2 \sqrt{t}$$ in order to get $$x \ n = \frac{x}{2 \sqrt{t}} \ \eta$$ but I'm also not sure how to continue from this. Sep 9, 2021 at 14:44
• For the less sophisticated of us, what is the ‘Feynman trick’? Sep 18, 2021 at 3:44
• It is a very powerful trick for calculating integrals by finding a function that has the derivative of the integrand. It is good described here Sep 18, 2021 at 18:11

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 $$\lim_{t\to\infty}\text{erf} \left(\frac{x}{2 \sqrt{t}}\right)=\text{erf}(0)=0\text{ for all }x>0$$
• @SahanManodya is it possible that there are some small mistakes in it? In line 2 when I don't get the factor 2: $$\frac{\partial}{\partial t} \frac{\sin(xn)}{xn}e^{-t n^2} \\ = \frac{\sin(xn)}{xn} e^{-t n^2} (-n^2) \\ = \frac{\sin(xn)}{x} e^{-t n^2} (-n)$$ Also the derivation of $$\text{erf} (\frac{x}{2 \sqrt{t}}) = \frac{x}{\pi} T$$ where T is what you wrote. Still the calculation works fine, I just have to adapt the solution a bit, or am I missing something? Sep 17, 2021 at 13:02