challenging calculus II problem and gamma function

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khauna

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I am having trouble with a problem for calculus II. I have tried working on it and gotten no where, I am just missing some little trick in the very beginning, I can feel it, anyway.

the problem is

integrate sinx/sinx+cosx (bounds are from 0 to pi/2)

My next troubles involve the gamma function:

gamma(x) = integral t^(x-1)e^-t dt bounds are 0 to infinity)
my assignment is to show that the function converges if x > 0
(hint: basically i have to find out how the integrand behaves at t approaching 0 and +infinity)

any advice help answers will be appreciated

PS (how do you format things to get them looking all math like, like getting integral signs and stuff, so its easier to read?)


thanks
 
For the first one:

\(\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{sin(x)}{sin(x)+cos(x)}dx\)

One way is perhaps let \(\displaystyle u=tan(x), \;\ x=tan^{-1}(u), \;\ dx=\frac{1}{u^{2}+1}du\)

Make the subs and we get:

\(\displaystyle \int\frac{u}{(u+1)(u^{2}+1)}du\)

\(\displaystyle =\frac{1}{2}\int\frac{u}{u^{2}+1}du+\frac{1}{2}\int\frac{1}{u^{2}+1}du-\frac{1}{2}\int\frac{1}{u+1}du\)

Now, you could tackle each of these separately. Can you continue?. They shouldn't be too bad.

Let me get back about gamma. I like gamma.
 
Wow okay I really didn't see that substitution. Yup I can continue that. Thanks
 
I have part of a project on gamma functions here that im working on. So far I have evaluated all of them correctly except that one i posted and now this one too. Im proud of myself so far with these gamma functions, 2 out of 8 I cant do. thats not too bad right. haha

but this one is causing me some grief:

gamma (n + 1/2)

i know gamma (1/2) = squareroot(x) so thats got to be apart of the answer...
 
Wow, some problems with a little meat on them. :D

I believe we may have to go at this with several cases. I have done this before a while back.

I will help get you started. OK?.

Gamma is improper because we can easily see the upper limit of integration is infinity. But if \(\displaystyle 0<x<1\), then

it is again improper because it has a singularity at x=0.

For \(\displaystyle 0<x<1\), we note that \(\displaystyle \int_{0}^{1}t^{p}dt\) converges if p>-1. Assume x>0

Next we can note that if t>0, then \(\displaystyle t^{x-1}e^{-t}<t^{x-1}\).

Therefore, gamma converges by the Comparison test and by the fact that \(\displaystyle \int_{0}^{1}t^{p}dt\) converges if p>-1.

Can you show that?.

Now, for t>1, we must show convergence for \(\displaystyle \int_{1}^{\infty}t^{x-1}e^{t}dt\).

Can you show this by the comparison test?. Try breaking it up into two cases: 1 to z and z to infinity. If you get stuck let me know.

We can set up three integrals and since they have all been shown to converge, then our proof is complete.

\(\displaystyle \int_{0}^{\infty}t^{x-1}e^{-t}dt=\underbrace{\int_{0}^{1}t^{x-1}e^{-t}dt}_{\text{assume x less than 0, comparison test}}+\overbrace{\int_{1}^{z}t^{x-1}e^{-t}dt}^{\text{regular}}+\underbrace{\int_{z}^{\infty}t^{x-1}e^{-t}dt}_{\text{comparison test}}\)
 
khauna said:
I have part of a project on gamma functions here that im working on. So far I have evaluated all of them correctly except that one i posted and now this one too. Im proud of myself so far with these gamma functions, 2 out of 8 I cant do. thats not too bad right. haha

but this one is causing me some grief:

gamma (n + 1/2)

i know gamma (1/2) = squareroot(x) so thats got to be apart of the answer...
Click to expand...

This may be tough, but can you show:

\(\displaystyle {\Gamma(n+\frac{1}{2})=\frac{1\cdot 3\cdot 5 \cdot\cdot\cdot (2n-1)}{2^{n}}\cdot\sqrt{\pi}=\frac{(2n)!}{4^{n}n!}\sqrt{\pi}\)
 
Hrmmm, not sure, ill see what I can do.

oh and thanks for the first gamma function one, that makes sense.
 
First one: Let u = sin(x)/[1+cos(x)], then sin(x) = 2u/(1+u^2), cos(x) = (1-u^2)/(1+u^2), and dx = 2du/(1+u^2).

Substitute into the definite integral, use partial fractiions, and then evaluate, should get Pi/4.

Second one I'll leave to the purists.
 
galactus, your post on the first problem is incorrect (see my post), however since you left yourself "wiggle room",

"One way is perhaps" <- perhaps is your wiggle, I won't get on your case.

Note: If one lets u = tan(x), when x = 0, tan(0) = 0, but when x = Pi/2, tan(Pi/2) is undefined, ergo one is turning a proper integral into an improper integral.
 
Yes, I know. I thought I could sneak that in. :oops: :wink:

Oh well, here's another way that works out well.

\(\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{sin(x)}{sin(x)+cos(x)}dx\)

Rewrite as \(\displaystyle \frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{2}}\frac{sin(x)}{sin(x+\frac{\pi}{4})}dx\)

Let \(\displaystyle u=x+\frac{\pi}{4}, \;\ du=dx, \;\ x=u-\frac{\pi}{4}\)

But \(\displaystyle \frac{sin(u-\frac{\pi}{4})}{sin(u)}=\frac{-cos(u+\frac{\pi}{4})}{sin(u)}=\frac{1}{\sqrt{2}}\frac{sin(u)-cos(u)}{sin(u)}=\frac{1}{\sqrt{2}}(1-cot(u))\)

Then we get:

\(\displaystyle \frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}du-\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}cot(u)du\)

\(\displaystyle =\boxed{\frac{\pi}{4}-0=\frac{\pi}{4}}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For the Gamma, here is another hint. Use the identities \(\displaystyle {\Gamma}(\frac{1}{2})=\sqrt{\pi} \;\ and \;\ {\Gamma}(n+1)=n{\Gamma}(n)\)
 
Excellent galactus, good show. I'll remember this one as it is better than my analysis (less protracted).
 
Thanks Dr. G for the accolades.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Continuation of Gamma problem

Here is a little more on the gamma problem. I looked at it and derived this, but I won't bother proving every identity. Perhaps you can do those.

As I said before, we can use \(\displaystyle {\Gamma}(1/2)=\sqrt{\pi}\) and perhaps \(\displaystyle p{\Gamma}(p)={\Gamma}(p+1)\), but perhaps not as well.

Knowing that the beta function can be expressed as \(\displaystyle B(p,q)=\frac{{\Gamma}(p){\Gamma}(q)}{{\Gamma}(p+q)}\)

We can write \(\displaystyle {\Gamma}(n+1/2)=\frac{{\Gamma}(p){\Gamma}(1/2)}{B(n,1/2)}\)...[1]

But \(\displaystyle B(n,1/2)=\frac{2^{2n}(n!)^{2}}{n(2n)!}\), (I will leave you prove this one).

Now, we get by subbing this into [1]:

\(\displaystyle \frac{(n-1)!\sqrt{\pi}}{\frac{2^{2n}(n!)^{2}}{n(2n)!}}=\boxed{\frac{(2n)!\sqrt{\pi}}{4^{n}n!}}\)
 
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