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I'm trying to evaluate the following integral:

$$ \int _0^{+\infty} \frac{x^m}{(a+bx^n)^p}$$

$a>0, b>0, n>0$

Could you please say if my reasoning is correct?

I thought that it was a good idea to use Chebyshev theorem on the integration of binomial differentials to evaluate the integral. Then I change the integral to the form: $$ \int _0^{+\infty} x^m(a+bx^n)^{-p}$$

However, the task gives no information regarding $p$ and whether $m, n \in \Bbb{Z} $.

So I need to look at 4 cases:

  1. $p \in \Bbb{Z}$

    Then use the substitution $x=t^r$ , $dx=rt^{r-1}$ where $r$ is common denominator of rational numbers $m$ and $n$.

  2. $\frac{m+1}{n} \in \Bbb{Z}$

    Then use the substitution $t=\sqrt[r]{a+bx^n}$ where $r$ is denominator of $p$.

  3. $\frac{m+1}{n}+p \in \Bbb{Z}$

    Then use the substitution $t=\sqrt[r]{\frac{a+bx^n}{x^n}}$ where $r$ is denominator of $p$.

  4. If all 3 previous cases are not applicable, then it's impossible to evaluate the integral.

Is my reasoning correct? I don't really know how to proceed further with these substitutions to evaluate the integral since there are no specific numbers here.

Any help is very much appreciated!

Super-kenguru
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  • For the antiderivative, I am afraid that you are facing an hypergeometric function (multiplied by a bunch of terms).. – Claude Leibovici Jun 12 '17 at 09:06
  • Cheating (which means using a CAS), there is a closed form expression for the integral from $0$ to $\infty$. – Claude Leibovici Jun 12 '17 at 09:14
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    If $a,b>0$ the $a$ and $b$ parameters are actually irrelevant since they can be removed through a suitable substitution. The question boils down to finding $$\int_{0}^{+\infty}\frac{x^m}{(1+x^n)^p},dx$$ and by setting $1+x^n = u$ this integral can be evaluated through Euler's Beta function, as $\frac{1}{n},B!\left(p-\frac{m+1}{n},\frac{m+1}{n}\right).$ – Jack D'Aurizio Jun 12 '17 at 17:01
  • Thank you, @JackD'Aurizio! Could you please give me a hint about how to use beta-function? – Super-kenguru Jun 12 '17 at 17:34
  • @Super-kenguru: as I said: perform the substitution $1+x^n=u$ and recall that $\int_{0}^{1}u^{a-1}(1-u)^{b-1},du = B(a,b) = \frac{\Gamma(a),\Gamma(b)}{\Gamma(a+b)}.$ – Jack D'Aurizio Jun 12 '17 at 17:52
  • @JackD'Aurizio, I've got it! Thank you so much! You're probably a genius! – Super-kenguru Jun 13 '17 at 20:27
  • For anyone who may be interested in a more detailed explanation for the Beta Function here - I addressed this here - https://math.stackexchange.com/questions/3057298/solving-used-real-based-methods-int-0x-fractk-lefttn-a-rightm-d/3059031#3059031 –  Jan 09 '19 at 00:21

1 Answers1

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$ \text { Let } b x^n=a \tan ^2 \theta \text {, then } x=\left(\frac{a}{b} \tan ^2 \theta\right)^{\frac{1}{n}} $ and $$ \begin{aligned} & I=\frac{2}{n}\left(\frac{a}{b}\right)^{\frac{1}{n}} \int_0^{\infty} \frac{(\frac{a}{b})^\frac{m}{n} \tan ^{\frac{2 m}{n}} \theta}{\sec ^2 \theta} \tan ^{\frac{2}{n}-1} \theta \sec ^2 \theta d \theta \\ & =\frac{2}{n}\left(\frac{a}{b}\right)^{\frac{m+1}{n}} \int_0^{\infty} \sin ^{2(\frac{m+1}{n})-1 }\theta \cos ^{2(p-\frac{m+1}{n})-1} \theta d \theta \\ & =\frac{1}{n}\left(\frac{a}{b}\right)^{\frac{m+1}{n}} B\left(\frac{m+1}{n}, p-\frac{m+1}{n}\right) \\ & =\frac{1}{n}\left(\frac{a}{b}\right)^{\frac{m+1}{n}} \frac{\Gamma\left(\frac{m+1}{n}\right) \Gamma\left(p-\frac{m+1}{n}\right)}{\Gamma(p)} \\ & \end{aligned} $$

Lai
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