Prove Convergence
- Thread starter flykilla
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We can do it this way by playing off of a known integral.
Knowing that \(\displaystyle \int_{0}^{\infty}sin(x^{2})dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}\)
Now, \(\displaystyle \int_{0}^{\infty}\frac{sin(x)}{\sqrt{x}}dx\)
Let \(\displaystyle u=\sqrt{x}, \;\ u=x^{2}, \;\ 2du=\frac{1}{\sqrt{x}}dx\)
Making the subs gives us \(\displaystyle \int_{0}^{\infty}\frac{sin(x)}{\sqrt{x}}dx=2\int_{0}^{\infty}sin(u^{2})du=\sqrt{\frac{\pi}{2}}\)
We can actually evaluate using the Gamma function or complex analysis, but that would be rather complicated.
Knowing that \(\displaystyle \int_{0}^{\infty}sin(x^{2})dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}\)
Now, \(\displaystyle \int_{0}^{\infty}\frac{sin(x)}{\sqrt{x}}dx\)
Let \(\displaystyle u=\sqrt{x}, \;\ u=x^{2}, \;\ 2du=\frac{1}{\sqrt{x}}dx\)
Making the subs gives us \(\displaystyle \int_{0}^{\infty}\frac{sin(x)}{\sqrt{x}}dx=2\int_{0}^{\infty}sin(u^{2})du=\sqrt{\frac{\pi}{2}}\)
We can actually evaluate using the Gamma function or complex analysis, but that would be rather complicated.
Though that is very nice and conclusive- is there any other way just to prove convergence, either using the Squeeze or Comparison Theorem or some sort of Alternating Series rule? Just wondering because these are the chapters and concepts most recently gone over in class, and they were all the ways I tried, and failed, to find the answer.
BigGlenntheHeavy
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\(\displaystyle galactus, \ if \ we \ use \ a \ power \ series \ for \ \int_{0}^{\infty}sin(x^{2})dx, \ we \ get \ \infty-\infty+\infty-\infty+...\)
\(\displaystyle Now, \ I \ know \ that \ \int_{0}^{\infty}sin(x^{2})dx \ = \ \frac{1}{2}{\sqrt \frac{\pi}{2}}\)
\(\displaystyle How \ is \ this \ derived? \ If \ it \ involves \ complicated \ analysis, \ forget \ it, \ however \ if \ it \ is \ something\)
\(\displaystyle \ less \ esoteric, \ I \ would \ be \ interested.\)
\(\displaystyle Now, \ I \ know \ that \ \int_{0}^{\infty}sin(x^{2})dx \ = \ \frac{1}{2}{\sqrt \frac{\pi}{2}}\)
\(\displaystyle How \ is \ this \ derived? \ If \ it \ involves \ complicated \ analysis, \ forget \ it, \ however \ if \ it \ is \ something\)
\(\displaystyle \ less \ esoteric, \ I \ would \ be \ interested.\)
Hey Glenn:
I worked on this back in undergrad when I was into doing tough integrals and the Gamma function.
To do it without CA, we can start with the Gamma function.
\(\displaystyle \frac{1}{x^{p}}=\frac{1}{{\Gamma}(p)}\int_{0}^{\infty}y^{p-1}e^{-xy}dy\)
Noting that \(\displaystyle {\Gamma}(\frac{1}{2})=\sqrt{\pi}\)
When p=1/2, \(\displaystyle \frac{1}{\sqrt{x}}=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{e^{-xy}}{\sqrt{y}}dy\)
We can rewrite \(\displaystyle \int_{0}^{\infty}\frac{sin(x)}{\sqrt{x}}dx=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-xy}sin(x)}{\sqrt{y}}dydx\)
We can proceed from here by switching integration limits, using sub, and what not.
Such as \(\displaystyle \frac{1}{1+y^{2}}=\int_{0}^{\infty}e^{-xy}sin(x)dx\)
From there we can write \(\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{y}(1+y^{2})}dy\)
If we keep going with subs and so forth, we can get it down to the solution.
I worked on this back in undergrad when I was into doing tough integrals and the Gamma function.
To do it without CA, we can start with the Gamma function.
\(\displaystyle \frac{1}{x^{p}}=\frac{1}{{\Gamma}(p)}\int_{0}^{\infty}y^{p-1}e^{-xy}dy\)
Noting that \(\displaystyle {\Gamma}(\frac{1}{2})=\sqrt{\pi}\)
When p=1/2, \(\displaystyle \frac{1}{\sqrt{x}}=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{e^{-xy}}{\sqrt{y}}dy\)
We can rewrite \(\displaystyle \int_{0}^{\infty}\frac{sin(x)}{\sqrt{x}}dx=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-xy}sin(x)}{\sqrt{y}}dydx\)
We can proceed from here by switching integration limits, using sub, and what not.
Such as \(\displaystyle \frac{1}{1+y^{2}}=\int_{0}^{\infty}e^{-xy}sin(x)dx\)
From there we can write \(\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{y}(1+y^{2})}dy\)
If we keep going with subs and so forth, we can get it down to the solution.
BigGlenntheHeavy
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Thanks, galactus, as I had a hunch the Gamma function came into play as it hs been a long time since I did any calculus analysis.
I reckon we wouldn't HAVE to use gamma.
I done it a while back and probably still have it written down here somewhere. It'll be lots of LaTex though.
I will have to get into the mood to do all that Latexing.
BUT, back to the problem at hand. The original query was just to show convergence. The poster requested that be done with some sort of test as opposed to the way I showed. Understandable.
Here, I found a link: http://mathforum.org/library/drmath/view/52063.html
It involves sin(x^2), which can easily be transformed into \(\displaystyle \frac{sin(x)}{\sqrt{x}}\) as shown previously.
When faced with a converging integral or limit, I always wanted to evaluate and find what it converged to besides just testing for convergence. That is how I got into doing some of these tough integrals.
I done it a while back and probably still have it written down here somewhere. It'll be lots of LaTex though.
I will have to get into the mood to do all that Latexing.
BUT, back to the problem at hand. The original query was just to show convergence. The poster requested that be done with some sort of test as opposed to the way I showed. Understandable.
Here, I found a link: http://mathforum.org/library/drmath/view/52063.html
It involves sin(x^2), which can easily be transformed into \(\displaystyle \frac{sin(x)}{\sqrt{x}}\) as shown previously.
When faced with a converging integral or limit, I always wanted to evaluate and find what it converged to besides just testing for convergence. That is how I got into doing some of these tough integrals.
Again, thank you for the help. My teacher is a very evil, though likeable, grad student who nabbed the problems from a test he is probably taking somewhere up high in the the Math degree- I believe he thinks he is either funny, or thinks we may learn something utterly failing at doing these problems. I did learn a lot- I will not be too distressed if I do not well on this one, he has stressed that if we do well on the in-class test later, a more reasonable test for us petty little calc 2 kids, we will be guaranteed a good grade in the class. I'm gonna go put some ice on my brain.