Solving Integral Problems: Limits and Double Integrals

[mansi]

i think i don't have adequate theory to solve these problems...

1.given that f(x) =cos(x) sin^k(x) / (1+x). calculate integral of this function wrt x between limits 0 and pi/2 . then find the it's limit as k tends to infinity...

2.let f(x)={e^(-ax)-e^(-bx)}/x, 0< a< b .let I be the integral of f wrt x between 0 and infinity. express I as a double integral...

 

[kleinwolf]

1) Are you meaning : \sin^k(x)\textrm{ or }\sin(x)^k ?

 

[mansi]

the former...

 

[dextercioby]

Do you have to compute these 2

F(k)=:\int_{0}^{\frac{\pi}{2}}\left[\cos x\left(\frac{\sin^{k}x}{1+x}\right)\right] dx

\lim_{k\rightarrow +\infty} F(k)

...?

Daniel.

 

[dextercioby]

Here's the answer for

F(1)=\frac {1}{2}\mbox{Si}\left( 2+\pi \right) \cos 2-\frac {1}{2}\mbox{Ci}\left( 2+\pi \right) \sin 2-\frac {1}{2}\mbox{Si}\left( 2\right) \cos 2+ \frac {1}{2}\mbox{Ci}\left( 2\right) \sin 2

F(2)= -\frac{1}{4}\mbox{Si}\left( \frac {3}{2}\pi +3\right) \sin 3-\frac{1}{4}\mbox{Ci}\left( \frac {3}{2}\pi +3\right) \cos 3+\frac{1}{4}\mbox{Si}\left( 1+\frac{1}{2}\pi \right) \sin 1
+\frac{1}{4}\mbox{Ci}\left( 1+\frac{1}{2}\pi \right) \cos 1+\frac{1}{4}\mbox{Si}\left( 3\right) \sin 3+\frac{1}{4}\mbox{Ci}\left( 3\right) \cos 3-\frac{1}{4}\mbox{Si}\left( 1\right) \sin 1-\frac{1}{4}\mbox{Ci}\left( 1\right) \cos 1

Daniel.

 

[dextercioby]

Here you can compute other values of F.I computed F(78).

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced#reply

Daniel.

 

[kleinwolf]

Well the first is really unsolvable I think...even the latter is hard...maybe a start like this :

\frac{sin(x)^kcos(x)}{1+x}=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^k\frac{e^{ix}+e^{-ix}}{2}\frac{1}{1+x}

= \frac{1}{2^{k+1}i^k}\sum_{n=0}^k\left(\begin{array}{c} k\\n\end{array}\right)(-1)^{n-k}\frac{e^{i(2n-k+1)x}}{1+x}+\frac{e^{i(2n-k-1)x}}{1+x}

Remain to calculate things of the form \int_0^{\frac{\pi}{2}}\frac{e^{imx}}{1+x}=\int_0^{\frac{\pi}{2}}e^{imx}\sum_{p=0}^\infty(-1)^p x^p


...well no sorry, i get lost and wrong, maybe some contour integral with residues in the complex plane...or a recursion relation over k

 

[dextercioby]

I linked you to a website which uses WebMathematica.I used my own antique Maple to come up with this (see attached file).

I can only assume that the limit is 0...

Daniel.

 

[shmoe]

1. Do you need to find the integral exactly? A bound will suffice if it's just the limit you're interested in, 1/(1+x) is bounded by 1 on [0..pi/2], cos and sin are non-negative there.

 

[Zurtex]

I've messed about in mathematica and got that:

\int_0^{\frac{\pi}{2}} \frac{\cos x \sin^{1000000000} x}{1 + x} dx = 0 \quad \text{(to 100 dp)}

Also if you look at the graph of the function as you increase k for either the whole function or just sin^k (x) it starts to look like the area does tend towards 0.

Perhaps this is about looking at the properties of the function, it would seem that:

\lim_{k \rightarrow \infty} \sin^k x = \left\{ \begin{array}{rcl} <br /> 1 &amp; \mbox{for} &amp; x = \pi \left(2n + \frac{1}{2} \right) \quad n \in \mathbb{Z}\\ <br /> \text{Undefined} &amp; \mbox{for} &amp; x = \pi \left(2n - \frac{1}{2} \right) \quad n \in \mathbb{Z} \\ <br /> 0 &amp; \mbox{for} &amp; \text{elsewhere} <br /> \end{array}\right. <br />

Now using that to look at the original function it's fairly easy to derive that:

\lim_{k \rightarrow \infty} \frac{\cos x \sin^{k} x}{1 + x} = 0 \quad \forall x \in \left[0,\frac{\pi}{2}\right]

Suddenly looks a lot easier to integrate to me

 

[mansi]

thanks for the help...
never mind the second one...i figured it out.

 

[kleinwolf]

Do you know why we always write sin(x)^k=sin^k(x) ?

This seems to me a notation abuse. Let's take k positive integer, then

sin(x)^k=sin(x)*sin(x)*...sin(x) k times

sin^k(x)=sin(sin(...sin(x)))...)) k times iterated

Or you have another notation for iteration n-times ?

 

[mansi]

sin^k(x)=(sinx)^k =sinx *sinx*...k times
and to express the second one let f(x)=sinx
then f^k(x)= (sin(sin(...sin(x)))...))

 

[Zurtex]

Actually it's not to confuse \sin (x^k) with (\sin x)^k as most people don't bother writing out the brackets.

But yeah notations don't seem to be entirely generalized.

 

[whozum]

klein, I've never seen the sin function self-iterated like you demonstrated,

sin^k(x) = sin(x)^k

The one on the RHS isn't used because of this one,
and they arent equalto sin(x^k)

 

[dextercioby]

Actually the argument of \sin x is never paranthesized,unless it gets really complicated...There's no point in writing \sin\left(x\right),but it is for \sin\left(x+y\right).

Daniel.

 

[Zurtex]

Actually the argument of \sin x is never paranthesized,unless it gets really complicated...There's no point in writing \sin\left(x\right),but it is for \sin\left(x+y\right).

Daniel.

But that's my point, can you see the difference between:

\sin {x^k}

And:

{\sin x}^k

Unless you look at the code?

 

[dextercioby]

I didn't look.Did u use "\displaystyle" ...?

Daniel.

EDIT:No,you didn't. I'm not trying to impose my own beliefs,just to stress out that writing

\sin x^{k}\neq \left(\sin x\right)^{k}\equiv \sin^{k} x

has a motivation.If someone else sees it and agrees on it,then I'm happy...

 

[Moo Of Doom]

But it does bother me that \sin^2{x}=(\sin{x})^2 but \sin^{-1}{x}\neq \frac{1}{\sin{x}}! Why use the same notation for these two concepts? Why does the k in sin^k(x) represent a functional power if k=-1, but not for any other k? Bah!

 

[whozum]

Wonder what sin^{-2}x would translate as?

 

[Zurtex]

Wonder what sin^{-2}x would translate as?

sin^{-2}x = \frac{1}{(\sin x)^2}

The notation may be odd but it's fairly common and well known.

 

[Moo Of Doom]

Hmm... on that note, what do you think \sin^0{x} is? 1 or x? I vote for 1.

 

[whozum]

Well technically only sin^{-1}x is the arcsin function, so how about

sin^{(-1)(0)}x

arcsin(x)^0 \ or \ 1^{-1}

 

[Zurtex]

Hmm... on that note, what do you think \sin^0{x} is? 1 or x? I vote for 1.

You'd be right, apart from when x = k\pi \quad k \in \mathbb{Z}. Look it's really simple in general:

\sin^a x = \left\{ \begin{array}{rcl} <br /> \arcsin x &amp; \mbox{for} &amp; x = -1\\ <br /> (\sin x)^a &amp; \mbox{for} &amp; \text{elsewhere} \\ <br /> \end{array}\right. <br />