# Evaluating an integral issue

• I
Hi,

I have the following integral that I want to evaluate:

$$\int_0^{\infty}y\,e^{-y\left[(z+1)(K-1)+1\right]}Ei\left(y_2(K-1)\right)\,dy$$

In the table of integrals there is a similar integral in the form

$$\int_0^{\infty}x^{v-1}\,e^{-\mu x}Ei\left(-\beta\,x\right)\,dx=-\frac{\Gamma(v)}{v(\mu+\beta)^v} 2F_1(1,\,v;\,v+1;\,\frac{\mu}{\mu+\beta})$$

and the conditions are ##|\arg{\beta}|<\pi##, ##Re\left\{\mu+\beta\right\}>0##, and ##Re\{v\}>0##. I think I meet all the conditions in my integral, but upon evaluating the result for ##z=2## and ##K=4##, it gives me ##\infty##!!. Why?

mathman
There may be a problem with the argument for Ei. You have y2(K-1) (y2???) while the formula has ##-\beta x##.. Yours seem to be positive.

There may be a problem with the argument for Ei. You have y2(K-1) (y2???) while the formula has ##-\beta x##.. Yours seem to be positive.

Oh, it is ##y## not ##y_2##. You are right, the argument of Ei(.) in my case is positive, but there is no condition that says that ##\beta## has to be positive. It says that ##\Re\{\mu+\beta\}## is positive, which is met.

mathman
I haven't work with the ##F_1## function, but looking at the expression, how did you get ##\infty##? The denominator = 98.

I haven't work with the ##F_1## function, but looking at the expression, how did you get ##\infty##? The denominator = 98.

I think the absolute value of the last argument of 2F1 function must be less than one. In my case, the last argument is greater than one!! But again, I am meeting the conditions of the integral, so, I why I get arguments that aren't right!

mathman