# 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, 11 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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, 11 months ago

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

- 2 years, 10 months ago

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

- 2 years, 10 months ago

I will post the answer on $$17$$th August.

- 2 years, 11 months ago