I need to solve the following integral $$\int_d^{\infty}\left(\frac{a}{bx-c}\right)^{p}x^f e^{-x}dx$$ where $a,b,c,d$ are all positive values and $p,f$ are positive integers. I have checked in the book "Integral series and products" by Ryzhik but I am unable to find an expression that matches with mine. I will be very thankful if somebody guides me how to solve the above integral. Thanks in advance.

  • $\begingroup$ @AlanTuring thanks for your comment. I have edited my question according to your comment. Can it be solved now? $\endgroup$ – Frank Moses Jan 31 '17 at 8:38
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    $\begingroup$ Oh! Sorry, I deleted my comment because I have read again the question and I noticed you wrote that $p$ and $e$ are integers. I thought I missed this line before... Sorry! Anyway, if this is the case then I'm going to think about this. $\endgroup$ – Turing Jan 31 '17 at 8:40
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    $\begingroup$ Just another question: two $e$ appear: are they both constants or is there an Euler number too between them? $\endgroup$ – Turing Jan 31 '17 at 8:40
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    $\begingroup$ @AlanTuring sorry. I changed it now. now there is only one $e$ and that is Euler number. $\endgroup$ – Frank Moses Jan 31 '17 at 8:43
  • $\begingroup$ @AlanTuring The expressions [Eq. 3.384.3-6] in the book "Integral series and products" are much similar to my expression but there are some differences which make my expression different. $\endgroup$ – Frank Moses Jan 31 '17 at 9:03

Partial Hint

Assuming all your parameters are positive (the fact they also be integers is another question that does not stress the following idea), you can think about integrating by parts, choosing for example

$$f'(x) = \left(\frac{a}{bx-c}\right)^p ~~~~~~~~~~~ f(x) = \frac{(c-b x) \left(\frac{a}{b x-c}\right)^p}{b (p-1)}$$

$$g(x) = x^f e^{-x} ~~~~~~~~~~~ g'(x) = e^{-x}x^{\ f-1}(f-x)$$

Hence you'd get

$$\frac{(c-b x) \left(\frac{a}{b x-c}\right)^p}{b (p-1)}x^f\ e^{-x}\bigg|_d^{+\infty} - \int_d^{+\infty} \frac{(c-b x) \left(\frac{a}{b x-c}\right)^p}{b (p-1)} e^{-x}x^{\ f-1}(f-x)\ \text{d}x$$

The first term gives you

$$\frac{e^{-d} d^f (c-b d) \left(\frac{a}{b d-c}\right)^p}{b (p-1)}$$

The second term, the integral, can be arranged to be written in this way:

$$\frac{a^p}{b(p-1)}\int_d^{+\infty} \frac{e^{-x} x^{\ f-1} (f-x)}{(bx-c)^{p-1}}\ \text{d}x$$

Which can be split into two pieces which are

$$f \frac{e^{-x}x^{\ f-1}}{(bx-c)^{p-1}} ~~~~~~~~~~~~~ (1)$$


$$\frac{e^{-x} x^{\ f}}{(bx-c)^{p-1}} ~~~~~~~~~~~~~ (2)$$

Which both have to be integrated.

Hence the "new" problem is to deal with those two integrals. I'll think more about. I'll add details later!

  • $\begingroup$ Thank you so much for your answer. I think now its much easier than before. $\endgroup$ – Frank Moses Jan 31 '17 at 12:00
  • $\begingroup$ Your answer is quite logical and easily comprehensible but I have one question. Please share your thoughts on following strategy. Suppose we use substitution $bx-c=u$ and solve the indefinite integral (without limits) in my first formula. We can solve it through the application of Binomial theorem. After getting the answer for indefinite integral we plug the limits? Will that answer be wrong? Please note that there are no other restrictions on the values of $a,b,c,d,f,p$ except those which are already mentioned in my original question. Thanks for your help. $\endgroup$ – Frank Moses Feb 2 '17 at 1:33

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