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# Integration Challenge!

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.

Note by Kartik Sharma
2 years, 1 month ago

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Master 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 $$\boxed{1}$$),

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

$\displaystyle = - \frac{1}{c} {\left(\frac{c}{a}\right)}^{\frac{s+1}{2}} \frac{\pi}{sin((s+1)\pi)} \frac{sin\left(s {cos}^{-1}\left(\frac{b}{2\sqrt{ac}}\right)\right)}{\sqrt{1 - \frac{{b}^{2}}{4ac}}}$ · 2 years, 1 month ago

I see that your last formula implies the divergence if the integral at $$s=0$$. How is that? · 2 years ago

Oh sorrry, I didn't see the $$s$$ behind the $$\cos ^{-1}$$ · 2 years ago

I will post the answer on $$17$$th August. · 2 years, 1 month ago