Disclaimer: This is a question about the mathematical concept, so I thought that it is appropriate to post my question here and not PSE
Hello MSE!
I was kind of annoyed of the derivatives in Schrodinger's equation. So I took the integral of this equation with respect to $t$ from both sides: $$i\hbar\int\frac{\partial}{\partial t} \psi (x,t)dt=\frac{-\hbar^2}{2m}\int\frac{\partial^2}{\partial x^2} \psi(x,t)dt$$ $$i\hbar\psi(x,t)=\frac{-\hbar^2}{2m}\int\frac{\partial^2}{\partial x^2} \psi(x,t)dt$$ $$\frac{-2im}{\hbar}\psi (x,t)=\int\frac{\partial^2}{\partial x^2}\psi(x,t)dt$$ The integral in the third line is reminiscent to differentiation under the integral sign except that there are no limits in the integration sign and we have the second derivative under it. I thought of adding limits to the integral: $$\int\frac{\partial^2}{\partial x^2}\psi(x,t)dt=\int^t_a\frac{\partial^2}{\partial x^2}\psi(x,t)dt$$ Where $\int\frac{\partial^2}{\partial x^2}\psi(x,a)dt$ then substituting $t=a$ gives $0$. The problem is, $t$ isn't a function of $x$, so I can't apply Leibniz's integral rule. Is there any other way to simplify this integral? And is my working correct?
Edit: Thanks to @KurtG and @whoisit, we can also write this as: $$\frac{2im}{\hbar}(\psi(x,a)-\psi(x,t))=\int^t_a\frac{\partial^2}{\partial x^2}\psi(x,s)ds+C$$ where $C$ is the constant of integration.