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.

  • $\begingroup$ Can you say something about $L$? Integer,real, complex...and the rest of parameters?. $\endgroup$ – popi May 1 '20 at 12:46
  • $\begingroup$ L is a positive intege(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. $\endgroup$ – Diagram May 1 '20 at 12:50

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

  • $\begingroup$ many thanks, I will try today to solve. But can you recommend ref book or anything how could I learned fraction decomp. such as that equations. $\endgroup$ – Diagram May 1 '20 at 22:43
  • $\begingroup$ Could you value my answer? $\endgroup$ – popi May 2 '20 at 0:26
  • $\begingroup$ I added an answer with a new post. Could you please look? $\endgroup$ – Diagram May 2 '20 at 6:41

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.

enter image description here

  • $\begingroup$ Both results seem similar, with different notation. $\endgroup$ – popi May 2 '20 at 12:54
  • $\begingroup$ What can you say about $\Omega_{n}$ I couldn't understand $\endgroup$ – Diagram May 2 '20 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.