The integrals $\int_0^\infty \frac{\mathrm{d}x}{(a+bx^c)^d(e+fx^g)^h}$ and $\int_0^\infty \frac{x^e}{(a+bx^c)^d}\mathrm{d}x$

Question

I am interested in improper integrals of rational functions. For example, I have found that $$\large{\int_0^\infty \frac{\mathrm{d}x}{(a+bx^c)^d}=\frac{\Gamma(\frac1c+1)\Gamma(d-\frac1c)}{\Gamma(d)a^{d-\frac1c}b^\frac1c}}$$ for $a,b,c,d\in\mathbb{C}$ selected such that the integral converges. Is there a way to generalise this further such as to find the value of $$\int_0^\infty \frac{\mathrm{d}x}{(a+bx^c)^d(e+fx^g)^h}\,\,\text{ or }\int_0^\infty \frac{x^e}{(a+bx^c)^d}\mathrm{d}x$$

Answer 1

Thanks to the comment Zacky made, I found the answer to the second integral by using the Beta function, in particular the integral definition $$B(x,y)=\int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}}\mathrm{d}t$$ giving the solution $$\large{\int_0^\infty \frac{x^e}{(a+bx^c)^d}\mathrm{d}x=\frac{a^{\frac{e+1}{c}}\Gamma(\frac{e+1}{c})\Gamma(d-\frac{e+1}{c})}{a^db^{\frac{e+1}{c}}c\,\Gamma(d)}}$$

Answer 2

Let us focus on the first integral. We assume that $a$,$b$,$e$ and $f$ are positive real numbers and that $c$,$g$,$d$ and $h$ are all positive integers bigger equal one. The starting point is the partial fraction decomposition: \begin{eqnarray} \frac{1}{a+b x^c} = \frac{1}{b c} \cdot \left( \frac{b}{a}\right)^{\frac{c-1}{c}} e^{-\imath \frac{\pi}{c} (c-1)} \sum\limits_{p=0}^{c-1} \frac{(-1)^{\frac{2 p}{c}}}{x - (\frac{a}{b})^{1/c} e^{\imath \pi \frac{2 p+1}{c}}} \end{eqnarray} Now writing another copy of the above with $(a,b,c)$ being replaced by $(e,f,g)$ respectively and then multiplying the original by the copy gives: \begin{eqnarray} &&\frac{1}{a+b x^c}\frac{1}{e+f x^g} = \frac{e^{-i \pi \left(-\frac{1}{c}-\frac{1}{g}+2\right)} \left(\frac{b}{a}\right)^{1-\frac{1}{c}} \left(\frac{f}{e}\right)^{1-\frac{1}{g}}}{b c f g} \cdot \\ && \sum\limits_{p=0}^{c-1} \sum\limits_{q=0}^{g-1} \left( \frac{1}{x-e^{\frac{i \pi (2 p+1)}{c}} \left(\frac{a}{b}\right)^{\frac{1}{c}}}-\frac{1}{x-e^{\frac{i \pi (2 q+1)}{g}} \left(\frac{e}{f}\right)^{\frac{1}{g}}} \right) \frac{(-1)^{\frac{2 p}{c}+\frac{2 q}{g}}}{e^{\frac{i \pi (2 p+1)}{c}} \left(\frac{a}{b}\right)^{\frac{1}{c}}-e^{\frac{i \pi (2 q+1)}{g}} \left(\frac{e}{f}\right)^{\frac{1}{g}}} \end{eqnarray} The expression above lends itself perfectly well for finding the anti-derivative with respect to $x$. Having done this we evaluate the the integral with respect to $x$ from zero to infinity and then we compute the $a$-derivative $(d-1)$ times and then the $e$-derivative $(h-1)$ times. Having done this we multiply the whole thing by $(-1)^{d+h}/((d-1)!(h-1)!)$. This gives: \begin{eqnarray} &&\int\limits_0^\infty \frac{1}{(a+b x^c)^d} \frac{1}{(e+f x^g)^h} dx=\frac{(-1)^{d+h}}{b c f g (d-1)! (h-1)!} e^{-i \pi \left(-\frac{1}{c}-\frac{1}{g}+2\right)}\cdot \\ && \frac{\partial^{d-1} }{\partial \xi^{d-1}} \frac{\partial^{h-1} }{\partial \eta^{h-1}} \left.\left[ \left(\frac{b}{\xi }\right)^{1-\frac{1}{c}} \left(\frac{f}{\eta }\right)^{1-\frac{1}{g}} \sum\limits_{p=0}^{c-1} \sum\limits_{q=0}^{g-1} \left( -\frac{\log \left(\frac{\xi }{b}\right)}{c}+\frac{i \pi (2 c q+c-2 g p-g)}{c g}+\frac{\log \left(\frac{\eta }{f}\right)}{g} \right) \frac{(-1)^{\frac{2 p}{c}+\frac{2 q}{g}}}{e^{\frac{i \pi (2 p+1)}{c}} \left(\frac{\xi }{b}\right)^{\frac{1}{c}}-e^{\frac{i \pi (2 q+1)}{g}} \left(\frac{\eta }{f}\right)^{\frac{1}{g}}} \right]\right|_{(\xi,\eta)=(a,e)} \end{eqnarray} It is clear that the result can be still simplified by using the chain rule of differentiation.

For[count = 1, count <= 50, count++,
  {a, b, e, f} = RandomReal[{0, 10}, 4, WorkingPrecision -> 50];
  {c, g, d, h} = RandomInteger[{1, 10}, 4];
  I1 = NIntegrate[1/(a + b x^c)^d 1/(e + f x^g)^h, {x, 0, Infinity}, 
    WorkingPrecision -> 30];
  I2 = 1/(b c ) 1/(f g ) (-1)^(
    d + h)/((d - 1)! (h - 1)!) (D[
       Exp[-I Pi (2 - 1/c - 1/g)] (b/xi)^(1 - 1/c) (f/eta)^(1 - 1/g)
         Sum[( -1/c Log[ xi/b ] + 1/g Log[eta/f ] + (
            I \[Pi] (c - g - 2 g p + 2 c q))/(c g)) (-1)^(
          2 p/c + 2 q/g)/((xi/b)^(1/c)
             Exp[(I Pi (2 p + 1))/c] - (eta/f)^(1/g)
             Exp[(I Pi (2 q + 1))/g]), {p, 0, c - 1}, {q, 0, 
          g - 1}], {xi, d - 1}, {eta, h - 1}] /. {xi :> a, eta :> e});
  If[Abs[I2/I1 - 1] > 10^(-3), 
   Print["mismatch..", {a, b, e, f, c, g, d, h}]; Break[]];
  PrintTemporary[count];
  ];