Find the general form for the following integral -

\[\displaystyle \large \int\limits_{0}^{\infty}{\frac{{x}^{s}}{a{x}^{2} + bx + c} dx}\]

for constants \(a, b, c, s\) such that the above integral converges. Find the condition for convergence as well.

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TopNewestMaster is here as many of you might have guessed.

First we will notice some facts about Chebyshev Polynomials of 2nd kind.

\(\displaystyle {U}_{n}(a) = \frac{sin((n+1){cos}^{-1}(a))}{sin({cos}^{-1}(a))}\)

And hence it can be seen that

\(\displaystyle \sum_{n=0}^{\infty}{{U}_{n}(a) {(-x)}^{n}} = \frac{1}{1 + 2ax + {x}^{2}}\)

\(\displaystyle \sum_{n=0}^{\infty}{{U}_{n}(a) \Gamma(n+1) \frac{{(-x)}^{n}}{n!}} = \frac{1}{1 + 2ax + {x}^{2}} ------\boxed{1}\)

Well, that may be all of the non-English we require, I guess. Back to the integral,

\(\displaystyle \int_{0}^{\infty}{\frac{{x}^{s}}{a{x}^{2} + bx + c} dx} = \frac{1}{c} \int_{0}^{\infty}{\frac{{x}^{s}}{\frac{a}{c}{x}^{2} + \frac{b}{c}x + 1} dx}\)

\(\displaystyle x \rightarrow \sqrt{\frac{a}{c}} x\)

\(\displaystyle \frac{1}{c} {\left(\frac{c}{a}\right)}^{\frac{s+1}{2}} \int_{0}^{\infty}{\frac{{x}^{s}}{{x}^{2} + \frac{b}{\sqrt{ac}}x + 1} dx}\)

Now, we can easily use Ramanujan Master Theorem(considering \(\boxed{1}\)),

\(\displaystyle = \frac{1}{c} {\left(\frac{c}{a}\right)}^{\frac{s+1}{2}} \Gamma(s+1)\Gamma(1-s-1) {U}_{-s-1}\left(\frac{b}{2\sqrt{ac}}\right)\)

\[\displaystyle = - \frac{1}{c} {\left(\frac{c}{a}\right)}^{\frac{s+1}{2}} \frac{\pi}{sin((s+1)\pi)} \frac{sin\left(s {cos}^{-1}\left(\frac{b}{2\sqrt{ac}}\right)\right)}{\sqrt{1 - \frac{{b}^{2}}{4ac}}}\] – Kartik Sharma · 1 year, 9 months ago

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– Hasan Kassim · 1 year, 9 months ago

I see that your last formula implies the divergence if the integral at \(s=0\). How is that?Log in to reply

– Hasan Kassim · 1 year, 9 months ago

Oh sorrry, I didn't see the \(s\) behind the \(\cos ^{-1} \)Log in to reply

I will post the answer on \(17\)th August. – Kartik Sharma · 1 year, 9 months ago

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