Solving an Integral with the Ei Function

[Pyrrhus]

\int \frac{1 - x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx

Here's what i have done

\int \frac{1 - x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx = \int \frac{1}{\frac{1}{y} + x \ln {y}} dx + \int \frac{- x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx

so

I_{1} = I_{2} + I_{3}

where

I_{1} = \int \frac{1 - x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx

I_{2} = \int \frac{1}{\frac{1}{y} + x \ln {y}} dx

I_{3} = \int \frac{- x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx

Now,

I_{2} = \int \frac{1}{\frac{1}{y} + x \ln {y}} dx

multiplying by \frac{\ln{y}}{\ln{y}}

I_{2} = \frac{\ln{y}}{\ln{y}} \int \frac{1}{\frac{1}{y} + x \ln {y}} dx

I_{2} = \frac{1}{\ln{y}} \int \frac{\ln{y}}{\frac{1}{y} + x \ln {y}} dx

Using substitution u = \frac{1}{y} + x \ln {y} and therefore du = \ln{y} dx

Now,

I_{2} = \frac{1}{\ln{y}} \int \frac{1}{u} du

I_{2} = \frac{1}{\ln{y}} \ln{|\frac{1}{y} + x \ln {y}|}

Any ideas about I_{3} or the whole integral?

 

[whozum]

multiplying by \frac{\ln{y}}{\ln{y}}

I_{2} = \frac{\ln{y}}{\ln{y}} \int \frac{1}{\frac{1}{y} + x \ln {y}} dx

I_{2} = \frac{1}{\ln{y}} \int \frac{\ln{y}}{\frac{1}{y} + x \ln {y}} dx

Using substitution u = \frac{1}{y} + x \ln {y} and therefore du = \ln{y} dx

I don't think that's fair game.

 

[dextercioby]

There you go

\int \frac{1-xe^{xy}}{\frac {1}{y}+x\ln y}dx=\allowbreak \frac{\ln \left( \frac {1}{y}+x\ln y\right) }{\ln y}-\frac {1}{y\ln y}e^{xy}-\frac {1}{y\ln^{2}y}e^{-\frac 1{\ln y}}\func{Ei}\left( 1,-xy-\frac 1{\ln y}\right) +C


Daniel.

 

[Pyrrhus]

Thanks Dexter , i guess my first integral was correct (I_{2}) but the second one (I_{3}) , whoa what a mess...

By the way, any idea for solving the I_{3} ?, i haven't been able to see anything... maybe i can rewrite it someway... i'll let you all know if i can.

Whozum? what do you mean? y is being kept as a constant...

 

[dextercioby]

You'll have to dig it somehow (make algebraic calculations and prossibly changes of integration variable) as to get it in the standard form of the Ei function.



Daniel.