How to Deal with this Integral

[Saladsamurai]

Homework Statement

So I have a problem in which I have arrived at the following equation (which according to my professor is correct). I am having some trouble with the math now:

$$\int_0^x \left [ 1 - \left ( \frac{\xi}{x} \right )^{3/4} \right ]^{-1/3} \frac{dT_w}{d\xi}\,d\xi = C \qquad(1)$$

Where C is a big lump of constants. I was then told to assume that T_w is given by a power series:

$$T_w (\xi) = \sum_0^\infty a_n\xi^n\qquad(2)$$

Plugging (2) back into (1) gives:

$$\int_0^x \left [ 1 - \left ( \frac{\xi}{x} \right )^{3/4} \right ]^{-1/3}\sum_{n=0}^\infty na_n\xi^{n-1}\,d\xi = C \qquad(3)$$

Somehow, I am supposed to get from (3) that T_w is a constant by solving for the a_n's.

I am not really sure how to handle (3) mathematically? Can anyone offer up the next step?

 

[tiny-tim]

Hi Saladsamurai!

… Where C is a big lump of constants. …

Somehow, I am supposed to get from (3) that T_w is a constant by solving for the a_n's.

I don't get it …

if T_w is constant, then its derivative, in (1), is zero, so C in (1) is 0.

 

[Saladsamurai]

Hi tiny tim! Yeah... that doesn't make sense! Just forget about that part for now. How would you go about solving (3) as it stands now for your a_n's.

 

[grey_earl]

As it stands, it's not correct. The derivative of a constant is zero, so your sum starts from n=1, and not from n=0. Now since the integral is linear, so can take the sum out of the integral and then integrate each term.

 

[tiny-tim]

Since C is constant, my inclination would be to d/dx both sides.

 

[Saladsamurai]

Since C is constant, my inclination would be to d/dx both sides.

That's totally what I said to my professor! But he completely ignored my suggestion If I d/dx it I will get:

$$\left [ 1 - \left ( \frac{\xi}{x} \right )^{3/4} \right ]^{-1/3}\sum_{n=0}^\infty na_n\xi^{n-1} = 0$$

which looks easier to handle.

 

[hunt_mat]

Even if you could do the integeral you would get something like f(x)=C, is this what he is after?

 

[grey_earl]

Just so you don't run into difficulties: your derivative is not correct, because you have x as a limit and inside the integral.
$$d/dx \int_0^x f(x, \xi) d\xi = f(x, x) + \int_0^x d/dx f(x, \xi) d\xi$$
And that complicates things quite a lot.

EDIT: Not to mention that f(x, x) diverges unless a_n = 0 for n >= 1. But that's no proof that T=const, because it's an integrable singularity, so the original integral exists. Maybe that's the reason your professor didn't want you to derive it with respect to x? (no pun intended)

 

[I like Serena]

$$\int_0^x \left [ 1 - \left ( \frac{\xi}{x} \right )^{3/4} \right ]^{-1/3} \frac{dT_w}{d\xi}\,d\xi = C \qquad(1)$$

An alternative approach is to make a substitution first.

Let's take $$u = 1 - (\frac \xi x)^{3/4}$$

So we'll get from (1):

$$\int_1^0 u^{-1/3} \frac{dT_w}{du} du = C$$

Now we'll substitute:

$$T_w (u) = \sum_0^\infty a_nu^n \qquad(2)$$

yielding

$$\sum^{\infty}_{n=1} n a_n \int_1^0 u^{-1/3} u^{n-1} du = C$$

which evaluates to:

$$\sum^{\infty}_{n=1} n a_n \frac {-1} {n - 1/3} = C$$

From this we can deduce that a_n must be zero starting from some N.
That is assuming that the power series for T_w, its derivative and the combined integral are convergent.

That's as far as I get. Useful?

 

[Saladsamurai]

Hi Saladsamurai!


I don't get it …

if T_w is constant, then its derivative, in (1), is zero, so C in (1) is 0.

Actually tiny-tim, according to my professor, this is exactly correct and is the point of the problem. C is actually, as I said, I lump of constants. By discovering that C is zero, it allows me to set my 'expanded version' of C equal to zero and solve for an unknown. That is, I can say that C = A - B = 0 and get some useful information out of it.

Just so you don't run into difficulties: your derivative is not correct, because you have x as a limit and inside the integral.
$$d/dx \int_0^x f(x, \xi) d\xi = f(x, x) + \int_0^x d/dx f(x, \xi) d\xi$$
And that complicates things quite a lot.

EDIT: Not to mention that f(x, x) diverges unless a_n = 0 for n >= 1. But that's no proof that T=const, because it's an integrable singularity, so the original integral exists. Maybe that's the reason your professor didn't want you to derive it with respect to x? (no pun intended)

Hi there grey_earl Thanks for responding. Your response has helped to point out yet another hole in my understanding of calculus. I will probably start a new thread on just how to take the derivative of that function since I have a lot of little questions about it. Maybe you will join me there!

An alternative approach is to make a substitution first.

Let's take $$u = 1 - (\frac \xi x)^{3/4}$$

So we'll get from (1):

$$\int_1^0 u^{-1/3} \frac{dT_w}{du} du = C$$

Now we'll substitute:

$$T_w (u) = \sum_0^\infty a_nu^n \qquad(2)$$

yielding

$$\sum^{\infty}_{n=1} n a_n \int_1^0 u^{-1/3} u^{n-1} du = C$$

which evaluates to:

$$\sum^{\infty}_{n=1} n a_n \frac {-1} {n - 1/3} = C$$

From this we can deduce that a_n must be zero starting from some N.
That is assuming that the power series for T_w, its derivative and the combined integral are convergent.

That's as far as I get. Useful?

Hi I like Serena! This is quite useful! I have another thought that is quite similar that I will post in a minute. But you have given me an idea

 

[Saladsamurai]

An alternative approach is to make a substitution first.

Let's take $$u = 1 - (\frac \xi x)^{3/4}$$

So we'll get from (1):

$$\int_1^0 u^{-1/3} \frac{dT_w}{du} du = C$$

Now we'll substitute:

$$T_w (u) = \sum_0^\infty a_nu^n \qquad(2)$$

yielding

$$\sum^{\infty}_{n=1} n a_n \int_1^0 u^{-1/3} u^{n-1} du = C$$

which evaluates to:

$$\sum^{\infty}_{n=1} n a_n \frac {-1} {n - 1/3} = C$$

From this we can deduce that a_n must be zero starting from some N.
That is assuming that the power series for T_w, its derivative and the combined integral are convergent.

That's as far as I get. Useful?

I like Serena. I am going through your substitution since mine is not working out. I am curious about a few things though. I'll start with one: since we defined $u = 1 - (\frac \xi x)^{3/4}$, it woud seem that assuming $T_w (u) = \sum_0^\infty a_nu^n$ is not the same as assuming $T_w (\xi) = \sum_0^\infty a_n\xi^n$, is it?

 

[Saladsamurai]

I have tried the following for now and have reached an impasse. I still have x's and xi's in my integral that I don't know what to so with:

 

[I like Serena]

I have tried the following for now and have reached an impasse. I still have x's and xi's in my integral that I don't know what to so with:

You didn't finish the substitution.
There's still a xi in the integral that needs to be replaced.

Furthermore, boundaries of your integral have changed to [0, 1], meaning x is no longer part of the integral, which means in turn that you can move all x out of the integral.

In other words.

From your definition of z follows:

$$\xi = x z^{\frac 4 3}$$

Substituting gives:

$$\frac 4 3 \sum n a_n \int_0^1 (1-z)^{\frac {-1} 3} (x z^{\frac 4 3})^n x z^{\frac 1 3} dz = \frac 4 3 \sum n a_n x^{n+1} \int_0^1 \frac {z^{\frac 4 3 n + \frac 1 3}} {(1 - z)^{\frac 1 3}} dz$$