Problem with differentiation under integral sign

Question

Original problem: I have a problem in which i need to evaluate the integral: $$ \int_1^\infty \dfrac{\sqrt{r^2-1}e^{-\alpha r}}{r} dr\, $$ I have tried to evaluate it taking the $\alpha$ derivative, here i give you all the steps i have done: $$ \int_1^\infty \dfrac{\sqrt{r^2-1}e^{-\alpha r}}{r} dr\,=-\int d\alpha \int_1^\infty\sqrt{r^2-1}e^{-\alpha r}=-\int d\alpha \dfrac{K_1(\alpha)}{\alpha}\\ =-\frac{1}{4} G^{2,1}_{0,1}\left(\frac{\alpha}{2},\frac{1}{2} \middle| \begin{array}{c} 1 \\ -1/2,1/2,0 \\ \end{array} \right) $$ The two last steps in the last equation were obtained using Mathematica, and have been checked numerically, however i have calculated numerically the first integral for a few different values for $\alpha$ and it doesn't match the values the MeijerG function gives.

So now i think i have two possibilities:

-There is something wrong in the first step, the differentiation under integral sign doesn't work and i don't know why this could be the case.

-It's a numerical problem. Mathematica could be giving me wrong values of the integral, but i think this can not be because i have been changing the precision and the method for the numerical integral, obtaining the same results.

I want to know what is happening, i have tried similar examples with the same method and i have obtained the good results. Of course, if you have another way of evaluating this integral, don't hesitate, i would like to know it, but also i would like to know what is wrong here.

EDIT:

Comparison between numerical solution vs MeijerG solution (using differentiation under integral sign) as $\alpha$ functions:

EDIT 2:

I have found that the difference between the numeric integral and the analytic solution obtained using differentiation under integral sign is exactly $\pi/2$ for all $\alpha$. Actually I have found that for an integral of the form:

$$ \int_z^\infty \dfrac{(r^2-z)^c e^{-\alpha r}}{r} dr\, $$ Being z a positive number and $0<c<1$. The differentiation under integral sign method gives the result up to a function f(z,c).

For example:

-For z=1 and c=1/2, this function is equal to $\pi/2$

-For z=1 and c=1/3 this function is equal to $\pi/\sqrt{3}$

-For c=1/2 this function is exactly $f(z,1/2)=-1+(1+\frac{\pi}{2})z$

With this i can continue with my calculations but, does anyone know why does this happen?

Do you know a way for obtaining this function f(z,c) analytically?

Answer 1

I have not checked your solution, this is a general answer.

You have your original integral as a function of $\alpha$:

$$ I(\alpha)=\int_1^\infty \dfrac{\sqrt{r^2-1}e^{-\alpha r}}{r} dr\, $$

You find a derivative of this function:

$$\frac{d I(\alpha)}{d \alpha}=f(\alpha)$$

Now you are solving a first order ordinary differential equation. Notice, that the derivative of a constant is zero, meaning:

$$\frac{d ( I(\alpha)+C_1)}{d \alpha}=f(\alpha)$$

$$I(\alpha)=\int f(\alpha) + C_1=g(\alpha) + C_1$$

This means that you get an arbitrary constant $C_1$ in your solution (which does not depend on $\alpha$), so you need an initial condition for your function to find this constant.

In the case you described the constant apparently $\pi/2$.

An initial condition would be any known value of $I(\alpha)$ for particular $\alpha$, like $I(1)$ or $I(1/2)$.

Then you can find the constant from the equation:

$$I(\alpha)=g(\alpha)+C_1$$