Partial Fraction Decomposition of $\int_{0}^{\infty} \frac{e^{-\frac{w}{s}}}{\left(mw+A\right) \left( mw+ B\right)^{L}}dw$

Question

I'm sorry the title not allow more than 150 characters, I couldn't put a full-length integral equation I tried to simplify the equation to decrease less than 150 char.

Here is the below full of the equation that I try to solve, How can I apply partial fraction decomposition of the following equation? I don't know how could I take integral this question.

$$ \left(\frac{h}{rs}\right)\left(\frac{h}{c}\right)^L\int_{0}^{\infty} \frac{e^{-\frac{w}{s}}}{\left(\frac{h}{r}+mw+\frac{\left( m-1\right) z}{k}\right) \left( \frac{h}{c}+mw+\frac{\left( m-1\right) z}{k}\right) ^{L}}dw $$

If this integrated can solve with other methods please advise me, and share ref book.

L is a positive integer(1,2,3 ... L) r, another parameters are positive decimal number( such as 1.2, 2.4,0.8 ). z is the variable of another function like $$$ f_{z}(z)$. So z must be stay as z.

Thanks for your help from now, Best Regards.

Answer 1

Calling $A=\frac{h}{r}+\frac{\left( m-1\right)z}{k}$ and $B=\frac{h}{c}+\frac{\left( m-1\right)z}{k}$ and $B-A=\frac{h}{c}-\frac{h}{r}=\frac{h}{c \,r}(r-c)$, your integral reads

$$\left(\frac{h}{rs}\right)\left(\frac{h}{c}\right)^L\int_{0}^{\infty} \frac{e^{-\frac{w}{s}}}{\left(mw+A\right) \left( mw+ B\right)^{L}}dw$$

A change of variable $w/s\longrightarrow y$ gives

$$\left(\frac{h}{r}\right)\left(\frac{h}{c}\right)^L\int_{0}^{\infty} \frac{e^{-y}}{\left(ms\,y+A\right) \left( ms\,y+ B\right)^{L}}dy\tag 1$$

Now, splitting fraction in elementary fractions

$$ \frac{1}{\left(ms\,y+A\right) \left( ms\,y+ B\right)^{L}}=\frac{1}{(B-A)^{L}}\left[ \frac{1}{A+ms\,y}-\sum _{k=1}^{L} \frac{(B-A)^{k-1}}{ (B+ms\, y)^{k}}\right]$$

Putting into $(1)$ and taking account that $$\int_0^{\infty } \frac{e^{-y}}{(C+m s\, y)^n} \, dy= \frac{e^{\frac{C}{m s}}}{(ms)^n} \,\Gamma \left(1-n,\frac{C}{m s}\right)$$ you have

$$\boxed{\left(\frac{h}{rs}\right)\left(\frac{h}{c}\right)^L\int_{0}^{\infty} \frac{e^{-\frac{w}{s}}}{\left(mw+A\right) \left( mw+ B\right)^{L}}dw= \\=\frac{h\,r^{L-1}}{(r-c)^{L}\,m\,s}\left[e^{\frac{A}{ms}}E_1\left(\frac{A}{ms}\right)-\sum _{k=1}^{L} e^{\frac{B}{ms}}\left(1-\frac{A}{B}\right)^k\,E_k\left(\frac{B}{ms}\right)\right]}$$

with $E_k(z)=z^{k-1}\Gamma(1-k,z)$ the Generalized Exponential Integral (see Exponential Integral)

P.S.: Be careful, there may be some error with so much notation

Answer 2

I found an article similar to my case but a little bit different. Here is the integrand.

$$ \left(\frac{I_{max}\sigma^2_{h}}{\sigma^2_{f_{sp}}\gamma IPR}\right)\left(\frac{I_{max}\sigma^2_{h}}{\gamma IR}\right)^{L_{R}}\int_{\frac{I_{max}}{Ps}}^{\infty} \frac{\exp\left(-w\left(\frac{I_{max}\sigma^2_{h}+\gamma \sigma^2_{f_{sp}}}{I_{max}\sigma^2_{h}\sigma^2_{f_{sp}}}\right)\right)}{\left(\frac{I_{max}}{\gamma IPR}+w\right)\left(\frac{I_{max}}{\gamma IR}+w\right)^{L_{R}}}dw $$

In the paper integral converted as follows using partial fraction decomposition: it is similar to your solution but I couldn't understand the differential term.

Here all of the parameters decimal or integer and positive. Only LR is a positive integer.