- #1

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## Main Question or Discussion Point

Hi,

I have the following integral that I want to evaluate:

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

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

[tex]\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})[/tex]

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?

I have the following integral that I want to evaluate:

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

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

[tex]\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})[/tex]

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?