Prove that \[ \int_{-\infty}^{\infty} \frac{e^{-(t-i\gamma)^2/2}}{\sqrt{2\pi}}\,dt \] is independent of the real parameter \(\gamma\).
Solution
For \(\gamma\in\RR\), define \[ I(\gamma) \coloneqq \frac1{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-(t-i\gamma)^2/2}\,dt. \]
1. The integral defining \(I(\gamma)\) converges absolutely for every\(\gamma\in\RR\).
For real \(t\) and \(\gamma\), \[ \abs{e^{-(t-i\gamma)^2/2}} = e^{-(t^2-\gamma^2)/2} = e^{\gamma^2/2}e^{-t^2/2}. \] The Gaussian on the right is integrable over \(\RR\).
2. The function \(I\) is differentiable and\[ I'(\gamma) = \frac{i}{\sqrt{2\pi}} \int_{-\infty}^{\infty} (t-i\gamma)e^{-(t-i\gamma)^2/2}\,dt. \]
Fix \(R>0\) and restrict temporarily to \(\abs{\gamma}\leq R\). The derivative of the integrand with respect to \(\gamma\) is \[ i(t-i\gamma)e^{-(t-i\gamma)^2/2}. \] Its modulus is at most \[ (\abs{t}+R)e^{R^2/2}e^{-t^2/2}, \] which is integrable over \(\RR\). Differentiation under the integral sign is therefore justified on \([-R,R]\). Since \(R\) is arbitrary, the displayed formula holds for every real \(\gamma\).
3. For every \(\gamma\in\RR\),\[ I'(\gamma)=0. \]
For fixed \(\gamma\), \[ \frac{d}{dt} e^{-(t-i\gamma)^2/2} = -(t-i\gamma)e^{-(t-i\gamma)^2/2}. \] Hence step 2 gives \[ I'(\gamma) = -\frac{i}{\sqrt{2\pi}} \left[ e^{-(t-i\gamma)^2/2} \right]_{t=-\infty}^{t=\infty}. \] By the modulus identity in step 1, the boundary values tend to zero as \(t\to\pm\infty\). Thus \(I'(\gamma)=0\).
4. The integral is independent of \(\gamma\).
Step 3 shows that the differentiable function \[ I:\RR\longrightarrow\CC \] has derivative zero everywhere. Hence \(I\) is constant on the connected interval \(\RR\).
5. Q.E.D.
Step 4 is exactly the required conclusion.