Exploring Multi-Variable Integral Limits with $\sin(t)/t$

[agnimusayoti]

Homework Statement

If $$u=\int_{x}^{(y-x)} \frac{sin t}{t} dt$$. Find $\frac {\partial{u}}{\partial{x}}. \frac {\partial{u}}{\partial{y}}, \frac {\partial{y}}{\partial{x}}$ at $x=\frac{\pi}{2}; y=\pi$

Relevant Equations

Leibniz Rule.

Because the limit of the integral is multi-variable, which is not explained at the ML Boas's example, I tried to start from the basic. First, I use:
$$\frac {dF}{dx}=f(x) \Rightarrow \int_a^b f(t) dt = F(b) - F(a)$$.
In my case now:
$$\int_{u(x)}^{v(x,y)} f(t) dt = F(v(x,y)) - F(u(x))$$
So,
$$\frac {d}{dx} \int_{u(x)}^{v(x,y)} f(t) dt = \frac {d}{dx} [F(v(x,y)) - F(u(x))]$$
$$\frac {d}{dx} \int_{u(x)}^{v(x,y)} f(t) dt = \frac {d}{dv} [F(v(x,y))]\frac {dv(x,y)}{dx} - \frac {d}{du} F(u(x)) \frac {du(x)}{dx}$$
$$\frac {d}{dx} \int_{u(x)}^{v(x,y)} f(t) dt = f(v) \frac {dv(x,y)}{dx} - f(u) \frac {du(x)}{dx}$$

where
$$dv(x,y)= \frac {\partial v}{\partial x} dx+\frac {\partial v}{\partial y} dy$$
So,
$$\frac {dv(x,y)}{dx}= \frac {\partial v}{\partial x} +\frac {\partial v}{\partial y} \frac {dy}{dx}$$
Now,
I put the detail on the question, where:
$$f(t) = \frac{\sin (t)}{t}$$
$v(x,y) = y - x/t$ then $dv = dy - dx$. So, $\frac {\partial v}{\partial x}=-1$ and $\frac {\partial v}{\partial y} = 1$.

Now, with this, can I solve the problem? Thanks...

 

[agnimusayoti]

I don't understand what is $\frac{\partial u}{\partial x}; \frac{\partial u}{\partial y}; \frac{\partial y}{\partial x}$

 

[Mark44]

I don't understand what is $\frac{\partial u}{\partial x}; \frac{\partial u}{\partial y}; \frac{\partial y}{\partial x}$

From your first post, the integral is $u=\int_{x}^{(y-x)} \frac{sin t}{t} dt$
u is a function of both x and y. The Fundamental Theorem of Calculus is your friend here. In part, it says that $$\frac d{dx} \left(\int_a^x f(t)dt \right) = f(x)$$

A difference with your problem is that the integral is a function of both x and y, rather than just x, so partial derivatives are called for.

A pair of useful formulas are
$\int_a^b f(t)dt = -\int_b^a f(t)dt$ -- I.e., switching the integration limits results in a sign change.
$\int_a^b f(t)dt = \int_a^c f(t)dt + \int_c^b f(t)dt$ -- this one is helpful in changing one of the integration limits in your problem to a constant.

Does Boas mention these concepts?

 

[agnimusayoti]

No. Maybe Boas think the student should've understood this concept in Calculus class. However, the second one is lost in my mind. Huah.

Then, what should I do after changing the integration limits? Thanks

 

[Mark44]

Maybe Boas think the student should've understood this concept in Calculus class.

If you've already had a course in which one topic was integration, then, yes, it's a reasonable assumption on the part of Mary Boas.

Here's from your first post:

If $u=\int_{x}^{(y-x)} \frac{sin t}{t} dt$. Find $\frac {\partial{u}}{\partial{x}}. \frac {\partial{u}}{\partial{y}}, \frac {\partial{y}}{\partial{x}}$ at $x=\frac{\pi}{2}; y=\pi$

From the above, u is a function of x and y, so u = g(x, y) = $\int_x^{y -x}\frac{\sin t}t dt$.

Split the integral into two integrals, like so:
$$u = \int_x^c \frac{\sin t}t dt + \int_c^{y - x} \frac{\sin t}t dt \\
= -\int_c^x \frac{\sin t}t dt + \int_c^{y - x} \frac{\sin t}t dt$$
Now take the partial with respect to x, using the Fundamental Theorem of Calculus (part of which I wrote in my earlier post).

This is the trick -- I'm afraid your work in post #1 was a wasted effort.

I don't understand what is $\frac{\partial u}{\partial x}; \frac{\partial u}{\partial y}; \frac{\partial y}{\partial x}$

Similar idea to find $\frac{\partial u}{\partial x}$. When you get those, we can look at $\frac{\partial y}{\partial x}$.

 

[agnimusayoti]

From your suggestion, I did this:
Take differential on both sides:
$$du=-d{\int_{c}^{x} \frac{\sin t}{dt}} + d{\int_{c}^{y-x} \frac{\sin t}{dt}}$$
$$du=-d{\int_{c}^{x} \frac{\sin t}{dt}} + d{\int_{c}^{y-x} \frac{\sin t}{dt}}$$
$$du=\frac{\sin {x}}{x} \frac {\partial {x}}{\partial x} dx +\frac{\sin (y-x)}{y-x} {\left[\frac {\partial \left(y-x \right)}{\partial x} dx +\frac {\partial {\left (y-x \right)}}{\partial y} dy \right]}$$
$$du=\frac{\sin {x}}{x} (1) dx +\frac{\sin (y-x)}{y-x} {\left[(-1) dx +(1) dy \right]}$$
$$du=\frac{\sin {x}}{x} (1) dx +\frac{\sin (y-x)}{y-x} {\left[(-1) dx +(1) dy \right]}$$
$$du=\left [\frac{\sin {x}}{x} - \frac{\sin (y-x)}{y-x} \right] dx +\left[\frac{\sin (y-x)}{y-x} \right] dy$$

So:
$$\frac {\partial u}{\partial x} = \frac{\sin {x}}{x} - \frac{\sin (y-x)}{y-x} $$
$$\frac {\partial u}{\partial y} = \frac{\sin (y-x)}{y-x} $$But, I don't understand how to find $\frac {\partial y}{\partial x}$?

 

[Mark44]

From your suggestion, I did this:
Take differential on both sides:
$$du=-d{\int_{c}^{x} \frac{\sin t}{dt}} + d{\int_{c}^{y-x} \frac{\sin t}{dt}}$$

This makes no sense. You can't have dt in the denominator of the integral, and you lost the t that was in the denominator.

Here's an example that uses the Fund. Thm. of Calculus.
$f(x) = \int_a^x \sin^2(t)dt$
$\Rightarrow f'(x) = \frac d{dx} \int_a^x \sin^2(t)dt = \sin^2(x)$
Notice that I don't ever attempt to actually perform the integration.

$$du=-d{\int_{c}^{x} \frac{\sin t}{dt}} + d{\int_{c}^{y-x} \frac{\sin t}{dt}}$$
$$du=\frac{\sin {x}}{x} \frac {\partial {x}}{\partial x} dx +\frac{\sin (y-x)}{y-x} {\left[\frac {\partial \left(y-x \right)}{\partial x} dx +\frac {\partial {\left (y-x \right)}}{\partial y} dy \right]}$$
$$du=\frac{\sin {x}}{x} (1) dx +\frac{\sin (y-x)}{y-x} {\left[(-1) dx +(1) dy \right]}$$
$$du=\frac{\sin {x}}{x} (1) dx +\frac{\sin (y-x)}{y-x} {\left[(-1) dx +(1) dy \right]}$$
$$du=\left [\frac{\sin {x}}{x} - \frac{\sin (y-x)}{y-x} \right] dx +\left[\frac{\sin (y-x)}{y-x} \right] dy$$

So:
$$\frac {\partial u}{\partial x} = \frac{\sin {x}}{x} - \frac{\sin (y-x)}{y-x} $$
$$\frac {\partial u}{\partial y} = \frac{\sin (y-x)}{y-x} $$

I didn't wade through all of your work above, because it was wrong in the very first line. What you have below might be the correct answer (is it shown in the textbook's solution?), but your work doesn't support it.

But, I don't understand how to find $\frac {\partial y}{\partial x}$?

If x and y are independent of each other then $\frac {\partial y}{\partial x} = 0$.

 

[agnimusayoti]

I think i was mistaken in typing. I mean sin t/t dt. Yes that is the solution andthe solution of $$\frac {\partial y}{\partial x}=-\frac{\frac {\partial u}{\partial x}}{\frac{\partial u}{\partial y}}$$. But i don't know how it can be. The solution is only numeric solution since the problem ask the numeric solution at $(\pi/2, \pi)$

 

[agnimusayoti]

Why i perform differential is because the book suggest to take differential. Thankss

 

[Mark44]

I think i was mistaken in typing. I mean sin t/t dt. Yes that is the solution andthe solution of $$\frac {\partial y}{\partial x}=-\frac{\frac {\partial u}{\partial x}}{\frac{\partial u}{\partial y}}$$. But i don't know how it can be. The solution is only numeric solution since the problem ask the numeric solution at $(\pi/2, \pi)$

I just noticed this near the bottom of post #1:

$v(x,y) = y - x/t$ then $dv = dy - dx$

v appears to be a function of three variables: x, y, and t, not just x and y.
Is t in the expression y - x/t in there by mistake? The first integral below has y - x as the upper limit of integration, not y - x/t.
These are from elsewhere in post #1.

If $u=\int_{x}^{(y-x)} \frac{sin t}{t} dt$.
<snip>
$\frac {d}{dx} \int_{u(x)}^{v(x,y)} f(t) dt$

Why i perform differential is because the book suggest to take differential. Thankss

I don't have the book, so I don't know what suggestions were given, but the problem is to find the partial derivatives of u (the integral) with respect to x and to y.

 

[agnimusayoti]

In my first post, I think I define so many functions and the same idea is used in your method to separate the limits of integration. By this, I can use fundamental theorem's of calculus since
$$\frac {d}{dx} \int_{x}^{c} \frac {\sin t}{t} dt = - \frac{ \sin x}{x}$$.

Nah, in #6 post, I did a mistake at earlier line, that is at 1st and 2nd line (actually they are the same). In that post, I write:

$$du=-d{\int_{c}^{x} \frac{\sin t}{dt}} + d{\int_{c}^{y-x} \frac{\sin t}{dt}}$$
$$du=-d{\int_{c}^{x} \frac{\sin t}{dt}} + d{\int_{c}^{y-x} \frac{\sin t}{dt}}$$

It supposed to be:
$$du=-d{\int_{c}^{x} \frac{\sin t}{t} dt} + d{\int_{c}^{y-x} \frac{\sin t}{t} dt}$$

Yeah, I'm afraid that was a mistake too. TT

I just noticed this near the bottom of post #1:

v appears to be a function of three variables: x, y, and t, not just x and y.
Is t in the expression y - x/t in there by mistake? The first integral below has y - x as the upper limit of integration, not y - x/t.
These are from elsewhere in post #1.

So sorry for this mistake.

But, how about the $\frac {\partial y}{\partial x}$? Could you explain why the solution seems satisfying this relation:

$$\frac {\partial y}{\partial x}=-\frac{\frac {\partial u}{\partial x}}{\frac{\partial u}{\partial y}}$$.

 

[Mark44]

But, how about the
$\frac {\partial y}{\partial x}$? Could you explain why the solution seems satisfying this relation:

$$\frac {\partial y}{\partial x}=-\frac{\frac {\partial u}{\partial x}}{\frac{\partial u}{\partial y}}$$.

No idea, at least based on the information you have presented here.
u is a function of x and y; i.e., we could write u = g(x, y), with g(x, y) being the integral in post #1.

There is no information shown here about a relationship between x and y, so we would conclude that $\frac {dy}{dx} = \frac{\partial y}{\partial x} = 0$.

Is there some information you have left out?

 

[agnimusayoti]

I think I have an idea.

$$du=\left [\frac{\sin {x}}{x} - \frac{\sin (y-x)}{y-x} \right] dx +\left[\frac{\sin (y-x)}{y-x} \right] dy$$

Maybe I can modify this to:
$$du- \left [\frac{\sin {x}}{x} - \frac{\sin (y-x)}{y-x} \right] dx = \left[\frac{\sin (y-x)}{y-x} \right] dy$$
Therefore I have y(u,x).

So,
$$\frac {\partial y}{\partial x}= \frac {- \left [\frac{\sin {x}}{x} - \frac{\sin (y-x)}{y-x} \right] }{\left[\frac{\sin (y-x)}{y-x} \right] } $$

Since, $$\left [\frac{\sin {x}}{x} - \frac{\sin (y-x)}{y-x} \right] = \frac {\partial u}{\partial x}$$ and $$\left[\frac{\sin (y-x)}{y-x} \right] = \frac {\partial u}{\partial y}$$ there fore,
$$\frac {\partial y}{\partial x}= - \frac {\frac {\partial u}{\partial x}}{\frac {\partial u}{\partial y}} $$

This idea just popped out when I see the equation again and again and again. Oh gee. But, thanks for your method, Mark44.

 

[Mark44]

Maybe I can modify this to:
$$du- \left [\frac{\sin {x}}{x} - \frac{\sin (y-x)}{y-x} \right] dx = \left[\frac{\sin (y-x)}{y-x} \right] dy$$
Therefore I have y(u,x).

Let me stop you right there. The original equation $u = \int_x^{y - x} \frac{\sin t} t~dt$.
The suggestion to calculate the total differential is a good one, so we have:
$$du = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy$$
Here the first partial derivative is the partial derivative of the integral with respect to x, and the second one is the partial derivative of the integral with respect to y.
To get each partial derivative, you have to use the Fundamental Theorem of Calculus, as I have been saying several times.

So, I get this for $\frac {\partial u}{\partial x}$:
$$\frac {\partial u}{\partial x} = \frac{ -\sin x} x - \frac{\sin(y - x)}{y - x} $$
This is different from what you have.
For $\frac {\partial u}{\partial y}$, I get the same as what you show below.

So,
$$\frac {\partial y}{\partial x}= \frac {- \left [\frac{\sin {x}}{x} - \frac{\sin (y-x)}{y-x} \right] }{\left[\frac{\sin (y-x)}{y-x} \right] } $$

This makes no sense algebraically. What happened to $du$ in your equation in the first quote?

Since, $$\left [\frac{\sin {x}}{x} - \frac{\sin (y-x)}{y-x} \right] = \frac {\partial u}{\partial x}$$ and $$\left[\frac{\sin (y-x)}{y-x} \right] = \frac {\partial u}{\partial y}$$ there fore,
$$\frac {\partial y}{\partial x}= - \frac {\frac {\partial u}{\partial x}}{\frac {\partial u}{\partial y}} $$

Are you getting this from a solution published in the Boas book? If so, I don't see that your work supports this conclusion.

 

[agnimusayoti]

Why negative?
No, the book just give final numeric solution.
I move dx from right side to the left, so I have y as a function of u and x.

But, it's interesting that with your work is very simple. Could you explain how you get $$\frac {\partial u}{\partial x}$$ instantly? Thanks.

 

[Mark44]

Why negative?
No, the book just give final numeric solution.
I move dx from right side to the left, so I have y as a function of u and x.

Like I said before, that algebra doesn't work.
Here's what you had before:

Maybe I can modify this to:
$$du- \left [\frac{\sin {x}}{x} - \frac{\sin (y-x)}{y-x} \right] dx = \left[\frac{\sin (y-x)}{y-x} \right] dy$$
Therefore I have y(u,x).

No, you don't have y as a function of u and x. What you have is an equation that involves x, y, dx, dy, and du.

So,
$$\frac {\partial y}{\partial x}= \frac {- \left [\frac{\sin {x}}{x} - \frac{\sin t}{t} \right] }{\left[\frac{\sin (y-x)}{y-x} \right] } $$

And somehow, u completely disappeared. That's why I said that the algebra doesn't work.

But, it's interesting that with your work is very simple. Could you explain how you get $$\frac {\partial u}{\partial x}$$ instantly? Thanks.

Edit: Fixed an error I had in one of the integrals.
Well, it wasn't instantly, but it wasn't too hard.
$\frac {\partial u}{\partial x} = \frac \partial {\partial x} \left(\int_x^c \frac{\sin t}{t} dt + \int_c^{y-x} \frac{\sin t}{t}dt\right)$
$= \frac \partial {\partial x} \left(-\int_c^x \frac{\sin t}{t} dt + \int_c^{y-x} \frac{\sin t}{t}dt\right)$
The first term simplifies to $-\frac{\sin x}{x}$. The other term simplifies to $\frac{\sin(y - x)}{y - x} \cdot (-1)$. That final factor of -1 comes from using the chain rule: $\frac \partial {\partial x} (y - x) = -1$.

 

[Delta2]

@Mark44 at your last post the second term inside the equality for $\frac{\partial u}{\partial x}$ should be $$\int_c^{y-x}\frac{\sin t}{t} dt$$ right? I believe you have a "serious" typo there (probably something that occurred from copy/paste thing))

 

[agnimusayoti]

Note: I (Mark44) had an error in the post quoted below. I have fixed the error in my post as well as in the quoted material below.

Like I said before, that algebra doesn't work.
Here's what you had before:

No, you don't have y as a function of u and x. What you have is an equation that involves x, y, dx, dy, and du.
And somehow, u completely disappeared. That's why I said that the algebra doesn't work.

Well, it wasn't instantly, but it wasn't too hard.
$\frac {\partial u}{\partial x} = \frac \partial {\partial x} \left(\int_x^c \frac{\sin t}{t} dt + \int_c^{y-x} \frac{\sin t}{t}dt\right)$
$= \frac \partial {\partial x} \left(-\int_c^x \frac{\sin t}{t} dt + \int_c^{y-x} \frac{\sin t}{t}dt\right)$
The first term simplifies to $-\frac{\sin x}{x}$. The other term simplifies to $\frac{\sin(y - x)}{y - x} \cdot (-1)$. That final factor of -1 comes from using the chain rule: $\frac \partial {\partial x} (y - x) = -1$.

Well youre right about this. But I am pretty sure about /partial y and x. Because implicitly there is a function of y(x,u)?

 

[Mark44]

@Mark44 at your last post the second term inside the equality for $\frac{\partial u}{\partial x}$ should be $$\int_c^{y-x}\frac{\sin t}{t} dt$$ right? I believe you have a "serious" typo there (probably something that occurred from copy/paste thing))

Yes, you are right. The integrand should be $\frac{\sin t}{t}$ throughout. I have edited my earlier post.
Not a copy/paste error, I was just thinking too far ahead of what I was typing.

 

[Mark44]

Note: I had an error in the post quoted below. I have fixed the error in my post as well as in the quoted material below.

Well youre right about this. But I am pretty sure about /partial y and x. Because implicitly there is a function of y(x,u)?

I think I understand what you're trying to say, but that's different from what you're actually saying. When you write y(x, u), you're saying that y is a function of x and u (which is a function of x and y).
For this problem, I think we can get $\frac{\partial y}{\partial x}$ from this relationship:
$$\frac{\partial y}{\partial x} = \frac{\frac{\partial u}{\partial x}}{\frac{\partial u}{\partial y}} $$

 

[agnimusayoti]

I think I understand what you're trying to say, but that's different from what you're actually saying. When you write y(x, u), you're saying that y is a function of x and u (which is a function of x and y).
For this problem, I think we can get $\frac{\partial y}{\partial x}$ from this relationship:
$$\frac{\partial y}{\partial x} = \frac{\frac{\partial u}{\partial x}}{\frac{\partial u}{\partial y}} $$

Negative isn't it? Because, if
$$du=\frac{\partial u}{\partial x} dx +\frac{\partial u}{\partial y} dy $$
then
$$du - \frac{\partial u}{\partial x} dx = \frac{\partial u}{\partial y} dy $$
To make total differential y in terms of dx and du, I divide both sides with $\frac{\partial u}{\partial y}$:
$$\frac{1}{\frac{\partial u}{\partial y}} du- \frac{\frac{\partial u}{\partial x}}{\frac{\partial u}{\partial y}} dx = dy $$
Implicitly:
$$dy = \frac{\partial y}{\partial x} dx + \frac{\partial y}{\partial u} du$$
Then,
$$ \frac{\partial y}{\partial x} = - \frac{\frac{\partial u}{\partial x}}{\frac{\partial u}{\partial y}} $$

 

[Mark44]

I will concede that your work seems correct. I had to try things out with a simpler example.
In my work, I'm going to write partial derivatives with subscripts, so $u_x$ means $\frac{\partial u}{\partial x}$.
Let $u = x^2y$
$u_x = 2xy$ and $u_y = x^2$.

From the equation, I can solve for y, with y being an implicit function of x and u.
$y = \frac u {x^2} = ux^{-2}$
So $y_u = x^{-2}$ and
$\frac{\partial y}{\partial x} = y_x = -2ux^{-3} = \frac{-2u}{x^3} = \frac{-2x^2y}{x^3} = \frac{-2y}{x}$

From the formula you derived,
$\frac{\partial y}{\partial x} = -\frac{u_x}{u_y} = -\frac{2xy}{x^2} = \frac{-2y} x$