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.


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


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?

  • 1
    $\begingroup$ Can you post the numerical values? $\endgroup$
    – science
    Feb 9, 2015 at 13:50
  • $\begingroup$ I give you a plot of the numerical values. Do you prefer a particular example? $\endgroup$
    – FMulder
    Feb 9, 2015 at 14:18

1 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:



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.