Solving Integral Problems: Limits and Double Integrals

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In summary: Daniel.In summary, the conversation mainly focuses on solving a given function and finding its limit as k tends to infinity. The first function given is f(x) = cos(x)sin^k(x)/(1+x) and the limit is found to be 0. The second function, f(x) = (e^(-ax) - e^(-bx))/x, is expressed as a double integral. The conversation also touches upon notations for sin function and its powers, such as sin(x)^k and sin^k(x). It is noted that sin(x)^k is not equal to (sin(x))^k and this notation may cause confusion. The conversation also briefly discusses the notation for inverse sin function, sin^-1
  • #1
mansi
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i think i don't have adequate theory to solve these problems... :frown:

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...
 
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  • #2
1) Are you meaning : [tex] \sin^k(x)\textrm{ or }\sin(x)^k [/tex] ?
 
  • #3
the former...
 
  • #4
Do you have to compute these 2

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

[tex] \lim_{k\rightarrow +\infty} F(k) [/tex]

...?

Daniel.
 
  • #5
Here's the answer for

[tex] 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 [/tex]

[tex] 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[/tex]
[tex]+\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 [/tex]

Daniel.
 
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  • #6
Here you can compute other values of F.I computed [itex]F(78) [/itex].

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

Daniel.
 
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  • #7
Well the first is really unsolvable I think...even the latter is hard...maybe a start like this :

[tex] \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} [/tex]

[tex] = \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} [/tex]

Remain to calculate things of the form [tex] \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 [/tex]


...well no sorry, i get lost and wrong, maybe some contour integral with residues in the complex plane...or a recursion relation over k
 
  • #8
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.
 
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  • #9
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.
 
  • #10
I've messed about in mathematica and got that:

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

Also if you look at the graph of the function as you increase k for either the whole function or just sink (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:

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

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

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

Suddenly looks a lot easier to integrate to me o:)
 
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  • #11
thanks for the help...
never mind the second one...i figured it out.
 
  • #12
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 ?
 
  • #13
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)))...))
 
  • #14
Actually it's not to confuse [itex]\sin (x^k)[/itex] with [itex](\sin x)^k[/itex] as most people don't bother writing out the brackets.

But yeah notations don't seem to be entirely generalized.
 
  • #15
klein, I've never seen the sin function self-iterated like you demonstrated,

[tex] sin^k(x) = sin(x)^k [/tex]

The one on the RHS isn't used because of this one,
and they arent equalto [tex] sin(x^k) [/tex]
 
  • #16
Actually the argument of [itex] \sin x [/itex] is never paranthesized,unless it gets really complicated...There's no point in writing [itex] \sin\left(x\right) [/itex],but it is for [itex] \sin\left(x+y\right) [/itex].

Daniel.
 
  • #17
dextercioby said:
Actually the argument of [itex] \sin x [/itex] is never paranthesized,unless it gets really complicated...There's no point in writing [itex] \sin\left(x\right) [/itex],but it is for [itex] \sin\left(x+y\right) [/itex].

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

[tex]\sin {x^k}[/tex]

And:

[tex]{\sin x}^k[/tex]

Unless you look at the code?
 
  • #18
I didn't look.Did u use "\displaystyle" ...?

Daniel.

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

[tex] \sin x^{k}\neq \left(\sin x\right)^{k}\equiv \sin^{k} x [/tex]

has a motivation.If someone else sees it and agrees on it,then I'm happy...
 
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  • #19
But it does bother me that [tex]\sin^2{x}=(\sin{x})^2[/tex] but [tex]\sin^{-1}{x}\neq \frac{1}{\sin{x}}[/tex]! 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!
 
  • #20
Wonder what [itex] sin^{-2}x [/itex] would translate as?
 
  • #21
whozum said:
Wonder what [itex] sin^{-2}x [/itex] would translate as?
[tex]sin^{-2}x = \frac{1}{(\sin x)^2}[/tex]

The notation may be odd but it's fairly common and well known.
 
  • #22
Hmm... on that note, what do you think [tex]\sin^0{x}[/tex] is? 1 or x? I vote for 1.
 
  • #23
Well technically only [itex] sin^{-1}x [/itex] is the arcsin function, so how about

[tex] sin^{(-1)(0)}x[/tex]

[tex] arcsin(x)^0 \ or \ 1^{-1} [/tex]
 
  • #24
Moo Of Doom said:
Hmm... on that note, what do you think [tex]\sin^0{x}[/tex] is? 1 or x? I vote for 1.
You'd be right, apart from when [itex]x = k\pi \quad k \in \mathbb{Z}[/itex]. Look it's really simple in general:

[tex]\sin^a x = \left\{ \begin{array}{rcl}
\arcsin x & \mbox{for} & x = -1\\
(\sin x)^a & \mbox{for} & \text{elsewhere} \\
\end{array}\right.
[/tex]
 

Related to Solving Integral Problems: Limits and Double Integrals

What is the difference between limits and double integrals?

Limits involve finding the value of a function as a variable approaches a specific value, while double integrals involve finding the area under a curve in two dimensions. In other words, limits focus on a single variable, while double integrals involve two variables.

What is the process for solving a limit problem?

To solve a limit problem, you must first determine the function and the variable that is approaching a specific value. Then, you can use algebraic manipulation, substitution, or graphing to evaluate the limit. Finally, check for any discontinuities or holes in the graph that may affect the limit.

How do you determine the limits of a double integral?

The limits of a double integral are determined by the bounds of the two variables being integrated. These bounds can be determined by the shape of the region being integrated over, the equations of the curves that define the region, and any restrictions on the variables.

What is the purpose of solving integral problems?

Solving integral problems allows us to find the area, volume, or other quantities under a curve or between two curves. This is useful in many areas of science, such as physics, engineering, and economics, where calculating these quantities can provide valuable information.

What are some common techniques for solving double integrals?

Some common techniques for solving double integrals include using the properties of integrals, changing the order of integration, and using substitution or trigonometric identities. Additionally, numerical integration methods, such as the trapezoidal rule or Simpson's rule, can be used for more complex integrals.

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