Derivative of integral $\int_0^{\infty} e^{-x \cosh t} dt$ I am given the integral $y = \int_0^{\infty}e^{-x\cosh t} dt$ and wish to show that this integral solves the modified Bessell equation: $x^2y'' + xy' -x^2y = 0.$
To do this I need to calculate the derivatives $y'$ and $y''$. I suspect integration by parts is necessary, but do not know how to continue. I have tried substituting $e^-x\cosh t$ and $\cosh t$ by their Taylor series and integrating term by term, but this does not seem to simplify things.
My question is: How do I calculating $y'$?
 A: You can check the equation as follows:
\begin{align*}
x^2(y'' - y)
&= x \int_{0}^{\infty} x (\cosh^2 t - 1) e^{-x \cosh t} \, dt \\
&= x \int_{0}^{\infty} x \sinh^2 t \, e^{-x \cosh t} \, dt \\
&= x \left[ -\sinh t \, e^{-x \cosh t} \right]_{0}^{\infty} + x \int_{0}^{\infty} \cosh t \, e^{-x\cosh t} \, dt \\
&= -xy'.
\end{align*}

@anthus, I agree that Leibniz integral rule (interchanging of integration and differentiation) is often not obvious, and is simply false in some cleverly designed examples.
In this case, however, we have a simple idea to circumvent such complication: think everything backward!
For example, suppose that you want to know that $y$ is differentiable with
$$ y'(x) = -\int_{0}^{\infty} \cosh t \, e^{-x \cosh t} \, dt. \tag{*} $$
To this end, let us work backward: That is, we will show that $y$ is an antiderivative of the right-hand side.
Denote the RHS of (*) as $y_1 (x)$. Clearly $y_1 (x)$ is a well-defined function. Now for any $0 < a < b$, Tonelli's theorem guarantees that we can interchange the order of integration and we have
\begin{align*}
\int_{a}^{b} y_1(x) \, dx
&= \int_{0}^{\infty} \int_{a}^{b} (-\cosh t) e^{-x \cosh t} \, dt dx \\
&= \int_{0}^{\infty} (e^{-b \cosh t} - e^{-a \cosh t}) \, dt \\
&= y(b) - y(a).
\end{align*}
This means that $y$ is an antiderivative of $y_1$, hence $y$ is differentiable and $y' = y_1$.
