If the expected value of a distribution is constant, it means its derivative with respect to the values it can take must be zero. I was wondering if there is a rigorous proof of the same.
Steps Tried $$ E\left(X\right)=\int_{-\infty}^{\infty}tf_{X}\left(t\right)dt $$ Using Leibniz Integral Rule and assuming the density is differentiable, $$ \frac{dE\left(X\right)}{dt}=\int_{-\infty}^{\infty}\left\{ f_{X}\left(t\right)+t\frac{\partial f_{X}\left(t\right)}{\partial t}\right\} dx+f_{X}\left(\infty\right)\frac{\partial\infty}{\partial t}-f_{X}\left(-\infty\right)\frac{\partial\left(-\infty\right)}{\partial t} $$ Using integration by parts, $$ \frac{dE\left(X\right)}{dt}=1+\left|tf_{X}\left(t\right)\right|_{-\infty}^{\infty}-1+f_{X}\left(\infty\right)\frac{\partial\infty}{\partial t}-f_{X}\left(-\infty\right)\frac{\partial\left(-\infty\right)}{\partial t} $$ Or directly using the definition of Integral as the Anti derivative, $$ \frac{dE\left(X\right)}{dt}=\left|tf_{X}\left(t\right)\right|_{-\infty}^{\infty} $$
How do we show these expressions to be zero. Can we take the derivative of infinity to be zero? (Is there a rigorous proof of the same?) And we also need the density of the function as the value tends to infinity?
I am just trying to understand and play around with probability and related concepts. Please let me know any mistakes in this line of thinking and any resources to further delve into these topics.