Problem with an Integral

1. Apr 21, 2005

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?

2. Apr 21, 2005

whozum

I dont think thats fair game.

3. Apr 21, 2005

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.

4. Apr 22, 2005

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...

Last edited: Apr 22, 2005
5. Apr 22, 2005

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.