0

I meet with a complex integral problem, given as follws: $$\int_0^\infty {\frac{{{x^p}}}{{{{\left( {x + a} \right)}^q}}} \cdot {{\left( {\frac{{x + b}}{{x + c}}} \right)}^n}dx,where{\text{ }}q > p + \frac{3}{2};a,b,c > 0;n = 0,1,2,3...} ,\infty $$

Chaoqing Tang
  • 71
  • What have you tried? You should start by considering specific cases and then generalising. For example, suppose $n=0$. Note that $\int_{C_{r}^{+}} f(z) \rightarrow 0$ as $r \rightarrow \infty$. You have the residue at $z = -a$ of $\binom{p}{q-1} \cdot (-a)^p$. So you know something, but there are signs of it being very ugly. When $p,q \notin \mathbb{Z}$, we have branch cuts. So even in the simplest of cases, the answer is somewhat intractable. Robert's answer below shows the complexity for low $n$. – Christopher K Apr 13 '14 at 05:51
  • thanks, i'll try your tips. – Chaoqing Tang Apr 14 '14 at 02:07

2 Answers2

0

Even for the case $n=1$ Maple gives a rather complicated result involving hypergeometric functions: $$ {\frac {{c}^{p}{a}^{-q+1}}{\Gamma \left( q \right) } \left( {c}^{-1-p }{a}^{1+p}{\mbox{$_2$F$_1$}(1,2+p;\,3-q+p;\,{\frac {a}{c}})}\Gamma \left( 2+p \right) \Gamma \left( -p+q-2 \right) +\pi \,{c}^{-q+1}{a} ^{q-1} \left( 1-{\frac {a}{c}} \right) ^{-q}\csc \left( \pi \,p-\pi \, q \right) \Gamma \left( q \right) \right) }+{\frac {{c}^{p-1}{a}^{-q +1}b}{\Gamma \left( q \right) } \left( \Gamma \left( 1+p \right) \Gamma \left( -1+q-p \right) {c}^{-p}{a}^{p} {\mbox{$_2$F$_1$}(1,1+p;\,p-q+2;\,{\frac {a}{c}})}-\pi \,{c}^{-q+1}{a} ^{q-1} \left( 1-{\frac {a}{c}} \right) ^{-q}\csc \left( \pi \,p-\pi \, q \right) \Gamma \left( q \right) \right) } $$

For larger $n$ it gets more complicated. I don't know if there is a single closed form that covers all $n$. However, you could take a generating function approach: if your integral is $J_n(a,b,c)$, then $$ \eqalign{\sum_{n=0}^\infty (-t)^n J_n(a,b,c) &= \int_0^\infty \dfrac{x^p}{(x+a)^q} \sum_{n=0}^\infty -t^n \left( \dfrac{x+b}{x+c}\right)^n\ dx \cr &= \int_0^\infty \dfrac{x^p}{(x+a)^q} \dfrac{x+c}{(1+t)(x+d)}\ dx = \dfrac{J_1(a,c,d)}{1+t}}$$ where $d = \dfrac{bt+c}{1+t}$.

Robert Israel
  • 448,999
  • I am unable to find my error since what I got for $n=1$ is $$\frac{\Gamma (p) a^{p-q} \Gamma (-p+q-1) (a p+b (-p+q-1))}{2 \Gamma (q)}$$ where $c$ does not appear. For $n=2$, I obtained $$\frac{a^{p-q-1} \left(a^2 \Gamma (p+1) \Gamma (-p+q-1)+b (2 a \Gamma (p) \Gamma (q-p)+b \Gamma (p-1) \Gamma (-p+q+1))\right)}{4 \Gamma (q)}$$ – Claude Leibovici Apr 13 '14 at 06:03
0

This is really more of an appendix to Robert's answer.

You have an integral dependent on six parameters, which we'll denote by $I_n(p,q;a,b,c)$. Some of the complexity of the integrand from the parameter dependence is entirely superficial and can be eliminated by transformations as simple as a change of scale. As you can see from the nasty cumbersome expression Robert posted, it is highly advantageous to get rid of as much of the superficial complexity as possible.

Substitute $x=ar$. Since $a\in\mathbb{R}^+$ by assumption, any variables and constants may be rescaled by $a$ for convenience. Now, $r=\frac{x}{a}$, $dx=a\,dr$, and

$$\int_{0}^{\infty}\frac{x^p}{(x+a)^q}\cdot\left(\frac{x+b}{x+c}\right)^ndx = \int_{0}^{\infty}\frac{(ar)^p}{(ar+a)^q}\cdot\left(\frac{ar+b}{ar+c}\right)^na\,dr\\ = a^{1+p-q}\int_{0}^{\infty}\frac{r^p}{(r+1)^q}\cdot\left(\frac{r+\beta}{r+\gamma}\right)^n\,dr$$

where we've defined $\beta:=\frac{b}{a}$, and $\gamma:=\frac{c}{a}$. So we see that the original six parameter integral can be reduced to a five parameter integral. To evaluate the integral for general $a$, it suffices to evaluate the $a=1$ case:

$$I_n(p,q;a,b,c)=a^{1+p-q}I_n(p,q;1,\beta,\gamma).$$

Other tricks you might consider include reduction formulas for $n$, or another substitution.

David H
  • 29,921