Partial fraction that contain special function

Question

How to apply partial fraction to the following equation:

$$ \frac{e^{\frac{(2c+5x)}{3x}} \mathop{E_{n}}\nolimits\!\left(x\right)}{(a+x)(b+x)} $$

Answer 1

No you can't do this: if you substitute your $A$ and $B$ into the RHS, then bring it all to a common denominator, you will find that you do not get the LHS. Actually if you think about it this is fairly obvious: the numerator on the RHS after combining terms will be a constant times $x$ plus a constant, and this is not what the numerator looks like on the LHS.

You can only do partial fractions for a rational function: that is, a polynomial divided by a polynomial. (Depending on exactly what you mean by partial fractions, it may also be necessary that the numerator has degree less than that of the denominator.)

You can do something a bit like partial fractions by taking out the non-polynomial terms, then putting them back in later: $$\eqalign{\frac{e^{\frac{(2c+5x)}{3x}}E_{n}(x)}{(a+x)(b+x)} &=e^{\frac{(2c+5x)}{3x}}E_{n}(x)\frac{1}{(a+x)(b+x)}\cr &=e^{\frac{(2c+5x)}{3x}}E_{n}(x) \Bigl(\frac{(b-a)^{-1}}{a+x}+\frac{(a-b)^{-1}}{b+x}\Bigr)\cr &=\frac{1}{b-a}\Bigl(\frac{e^{\frac{(2c+5x)}{3x}}E_{n}(x)}{a+x} -\frac{e^{\frac{(2c+5x)}{3x}}E_{n}(x)}{b+x}\Bigr)\ .\cr}$$ But if you are looking at doing say an integral, this is probably not much help.