# Evaluating an integral issue

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## Main Question or Discussion Point

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.