How Do You Solve the Integral of Sin^2008(x) from 0 to Infinity?

[siaosi]

Completely stuck with this:

$$\int_0^\infty sin^{2008} x dx$$


Any help on how to tackle this please?

 

[Hurkyl]

What have you tried? (And are you sure you wrote it correctly?)

 

[siaosi]

Thanks for you prompt reply,

Yes it's written correctly. I've tried a formula that I found in a book, that says it is:

1*3*5*...*(n-1)/2*4*6*...*(n) * pi

But it just can't be as simple as that... I'm sure this is a tricky question.

 

[Defennder]

Hmm, I have to say that doesn't seem particular plausible. Think of the area under the graph of $$\sin^{2008} x$$ from 0 to $$\infty$$. Did you mistype the integral limits?

 

[siaosi]

Ok, I have re-checked it, I had it wrong, it is from 0 to pi. Sorry guys.
$$\int_0^\pi sin^{2008} x dx$$Still buffed though!

 

[snipez90]

Hmm well you could apply the usual methods for starting these integrals. Consider breaking down the integral into two (one with limits of integration of 0 to pi/2 and another of pi/2 to pi). Then consider the symmetry of sin(x) to realize that these two integrals are equal. Then try a substitution involving the identity cos(x) = sin(pi/2 - x).

But then again in this case it might complicate the problem. Is there an easy way to factor sin²⁰⁰⁸(x) + cos²⁰⁰⁸(x)

 

[Pere Callahan]

I would suggest you try integration by parts to derive a recursion formula for

$$I_n=\int_0^\pi{dx \sin^nx}$$Then you can solve this recursion formula and plug in n=2008.

 

[snipez90]

I would suggest you try integration by parts to derive a recursion formula for

$$I_n=\int_0^\pi{dx \sin^nx}$$


Then you can solve this recursion formula and plug in n=2008.

I have seen that recursion formula before and I can tell you without remembering the specifics that 2008 is a high enough number that this approach will probably be not very pleasant.

 

[Pere Callahan]

I have seen that recursion formula before and I can tell you without remembering the specifics that 2008 is a high enough number that this approach will probably be not very pleasant.

I can tell you that the result is not very pleasant

 

[snipez90]

Haha, well I didn't mean to be rude, since my approach seemed to get no where. I forgot that $$a^n + b^n$$ cannot really be factored readily.

 

[tiny-tim]

$$(2n-1)\int_0^{\pi}\,sin^{2n}x\,dx$$

$$=\ (2n-1)\int_0^{\pi}\,sin^{2n-2}x\,dx\ -\ \int_0^{\pi}\,(2n-1)\,sin^{2n-2}x\,cos^2xdx$$

$$=\ (2n-1)\int_0^{\pi}\,sin^{2n-2}x\,dx\ -\ [sin^{2n-1}x\,cosx]_0^{\pi}\ -\ \int_0^{\pi}\,sin^{2n}x\,dx$$

So $$\int_0^{\pi}\,sin^{2n}x\,dx\ =\ \frac{2n-1}{2n}\,\int_0^{\pi}\,sin^{2n-2}x\,dx$$ ​

 

[Pere Callahan]

Yes, and the same works if you take n instead of 2n.

 

[tiny-tim]

probability …

Is $$\frac{1}{\pi}\,\int_0^\pi sin^{2008} x dx$$

the probability of tossing a coin 2008 times and getting exactly 1004 heads?

If so, why … ? ​

($$\int_0^\pi sin^m x dx$$ is rational for odd m, and a rational multiple of π for even m)​

 

[NoMoreExams]

How can it be? Tossing a coin is a discrete even -> discrete probability distribution function but integrals are analagous to continuous ones. Or am I misunderstanding your question tiny-tim?

 

[Pere Callahan]

Well, maybe, Tiny-Tim just observed that the two quantities - the integral on the one hand, and the describes probability - are the same.

The probability of having k heads in 2k tosses is

$$p_{2k}=\left(\stackrel{2k}{k}\right)\left(\frac{1}{2}\right)^{2k}=\frac{(2k!)}{k!k!}\left(\frac{1}{2}\right)^{2k}$$

On the other hand, from the recursion formula described above one finds (using $I_0=\pi$)

$$\frac{1}{\pi}I_{2k}=\frac{(2k-1)!}{(2k)!}$$

Now one can use the identites $(2k)!=2^k k!$ and $(2k-1)!=\frac{(2k)!}{2^k k!}$ to find

$$\frac{1}{\pi}I_{2k} = \frac{(2k)!}{2^kk!2^kk!}=p_{2k}$$

If the question were not about 2008 but some odd number, something similar would probably be true.

 

[tiny-tim]

$$\frac{1}{\pi}I_{2k} = \frac{(2k)!}{2^kk!2^kk!}=p_{2k}$$

Yes … but why?

what does it all mean? ​

 

[Pere Callahan]

I don't know ...

 

[siaosi]

Wow, thanks for all the answers!

$$(2n-1)\int_0^{\pi}\,sin^{2n}x\,dx$$

$$=\ (2n-1)\int_0^{\pi}\,sin^{2n-2}x\,dx\ -\ \int_0^{\pi}\,(2n-1)\,sin^{2n-2}x\,cos^2xdx$$

$$=\ (2n-1)\int_0^{\pi}\,sin^{2n-2}x\,dx\ -\ [sin^{2n-1}x\,cosx]_0^{\pi}\ -\ \int_0^{\pi}\,sin^{2n}x\,dx$$

So $$\int_0^{\pi}\,sin^{2n}x\,dx\ =\ \frac{2n-1}{2n}\,\int_0^{\pi}\,sin^{2n-2}x\,dx$$ ​

I have done my research and from various books I have found the same answer as this. This is all fine, but I have a loooooooong way to reach the number 2008... So I am still confused.

 

[Pere Callahan]

See posts #7 and #15.

 

[tiny-tim]

I have done my research and from various books I have found the same answer as this. This is all fine, but I have a loooooooong way to reach the number 2008... So I am still confused.

Hi siaosi!

The best thing to do, if you're confused by complicated equations, is to give things short names so that equatins look really simple.

In this case, define P_2n = ∫₀^π sin²ⁿx dx.

Then that equation was P_2n =

P_2(n-1) (2n-1)/2n,​

which you can then see is

P_2(n-2) (2n-1)(2n-3)/2n(2n-2),​

and so on … until you reach P₀, which you can easily work out!

Now can you see that there's a pattern … so you don't have to worry about making thousands of calculations!