Critical Points of a Parameter Dependent Integral

[Peter Alexander]

1. The problem statement, all variables, and given/known data
Find and categorize extremes of the following function: $$F(y)=\int_{y}^{y^{2}}\frac{1}{\ln^{2}x}dx$$ for $y>1$.

Homework Equations

$$\frac{d}{dx}\int_{a}^{b}f(x,y)dy=\int_{a}^{b}\frac{\partial}{\partial x}\left(f(x,y)\right)dy$$ and $$\frac{d}{dy}\int_{\alpha(y)}^{\beta(y)}f(x,y)dx=\int_{\alpha(y)}^{\beta(y)}\frac{\partial f(x,y)}{\partial y}dx+\beta'(y)\cdot f\left(\beta(y),y\right)-\alpha'(y)\cdot f\left(\alpha(y),y\right)
$$

The Attempt at a Solution

Applying the formula for derivative of a parameter dependent integral gives the following result:
$$
\frac{d}{dy}\int_{y}^{y^{2}}\frac{1}{\ln^{2}x}dx=\int_{y}^{y^{2}}\frac{\partial}{\partial y}\left(\frac{1}{\ln^{2}x}\right)dx=0
$$
but since boundaries of integration are parameter dependent, the other equation should be implemented
$$
\frac{d}{dy}\int_{\alpha(y)}^{\beta(y)}f(x,y)dx=\int_{\alpha(y)}^{\beta(y)}\frac{\partial f(x,y)}{\partial y}dx+\beta'(y)\cdot f\left(\beta(y),y\right)-\alpha'(y)\cdot f\left(\alpha(y),y\right)
$$
or in this case
$$
\frac{d}{dy}\int_{y}^{y^{2}}\frac{1}{\ln^{2}x}dx=0+\frac{2y}{\ln^{2}\left(y^{2}\right)}-\frac{y}{\ln^{2}\left(y\right)}=0
$$

This is true if, and only if, $y=0$.
--------------------------------------------
I've solved the problem up to this point, hitting a brick wall. Because no other solution besides $y=0$ is available, yet task explicitly says that $y\text{ must be }>0$, either something is wrong with my computation or this problem is unsolvable.

PS: this is my first post, so if you have any comment on the styling and/or content, please let me know so I can improve my questions in the future. Also, if anyone is willing to help me, I'd like to thank you in advance.

 

[Orodruin]

but since boundaries of integration are parameter dependent, the other equation should be implemented
$$
\frac{d}{dy}\int_{\alpha(y)}^{\beta(y)}f(x,y)dx=\int_{\alpha(y)}^{\beta(y)}\frac{\partial f(x,y)}{\partial y}dx+\beta'(y)\cdot f\left(\beta(y),y\right)-\alpha'(y)\cdot f\left(\alpha(y),y\right)
$$
or in this case
$$
\frac{d}{dy}\int_{y}^{y^{2}}\frac{1}{\ln^{2}x}dx=0+\frac{2y}{\ln^{2}\left(y^{2}\right)}-\frac{y}{\ln^{2}\left(y\right)}=0
$$

You have done the derivative of the lower bound incorrectly.

 

[Mark44]

1. The problem statement, all variables, and given/known data
Find and categorize extremes of the following function: $$F(y)=\int_{y}^{y^{2}}\frac{1}{\ln^{2}x}dx$$ for $y>1$.

Homework Equations

$$\frac{d}{dx}\int_{a}^{b}f(x,y)dy=\int_{a}^{b}\frac{\partial}{\partial x}\left(f(x,y)\right)dy$$ and $$\frac{d}{dy}\int_{\alpha(y)}^{\beta(y)}f(x,y)dx=\int_{\alpha(y)}^{\beta(y)}\frac{\partial f(x,y)}{\partial y}dx+\beta'(y)\cdot f\left(\beta(y),y\right)-\alpha'(y)\cdot f\left(\alpha(y),y\right)
$$

You don't need partial derivatives here. You can assume that y is a function of a single variable: x; i.e., y = f(x).

The Attempt at a Solution

Applying the formula for derivative of a parameter dependent integral gives the following result:
$$
\frac{d}{dy}\int_{y}^{y^{2}}\frac{1}{\ln^{2}x}dx=\int_{y}^{y^{2}}\frac{\partial}{\partial y}\left(\frac{1}{\ln^{2}x}\right)dx=0
$$

Split the integral into two parts: $F(y) = \int_y^a \ln^2(x) dx + \int_a^{y^2} \ln^2(x)dx$.
Now use the Fund. Thm. of Calculus to find F'(y). You will need to use the chain rule, since one of the limits of integration represents a function of y, which itself is a function of x.

but since boundaries of integration are parameter dependent, the other equation should be implemented
$$
\frac{d}{dy}\int_{\alpha(y)}^{\beta(y)}f(x,y)dx=\int_{\alpha(y)}^{\beta(y)}\frac{\partial f(x,y)}{\partial y}dx+\beta'(y)\cdot f\left(\beta(y),y\right)-\alpha'(y)\cdot f\left(\alpha(y),y\right)
$$
or in this case
$$
\frac{d}{dy}\int_{y}^{y^{2}}\frac{1}{\ln^{2}x}dx=0+\frac{2y}{\ln^{2}\left(y^{2}\right)}-\frac{y}{\ln^{2}\left(y\right)}=0
$$

This is true if, and only if, $y=0$.
--------------------------------------------
I've solved the problem up to this point, hitting a brick wall. Because no other solution besides $y=0$ is available, yet task explicitly says that $y\text{ must be }>0$, either something is wrong with my computation or this problem is unsolvable.

PS: this is my first post, so if you have any comment on the styling and/or content, please let me know so I can improve my questions in the future. Also, if anyone is willing to help me, I'd like to thank you in advance.

 

[Orodruin]

You don't need partial derivatives here. You can assume that y is a function of a single variable: x; i.e., y = f(x).

This is not correct, y and x are independent. In the expression you quoted, $y$ is the variable that the entire expression depends on and $x$ is the integration variable. There is no a priori relationship between them.

Split the integral into two parts: $F(y) = \int_y^a \ln^2(x) dx + \int_a^{y^2} \ln^2(x)dx$.
Now use the Fund. Thm. of Calculus to find F'(y). You will need to use the chain rule, since one of the limits of integration represents a function of y, which itself is a function of x.

But he did this already, all that went wrong was the differentiation of one of the boundaries.

 

[Mark44]

This is not correct, y and x are independent. In the expression you quoted, $y$ is the variable that the entire expression depends on and $x$ is the integration variable. There is no a priori relationship between them.

I'm going on the basis of the original problem statement, which says nothing about x and y being independent.

1. The problem statement, all variables, and given/known data
Find and categorize extremes of the following function: $$F(y)=\int_{y}^{y^{2}}\frac{1}{\ln^{2}x}dx$$ for $y>1$.

For this problem $F'(y) = \frac d{dy} \left( \int_{y}^{y^{2}}\frac{1}{\ln^{2}x}dx\right)$, and my approach would be exactly as I said in post #3.

 

[Mark44]

$\frac{d}{dy}\int_{y}^{y^{2}}\frac{1}{\ln^{2}x}dx=0+\frac{2y}{\ln^{2}\left(y^{2}\right)}-\frac{y}{\ln^{2}\left(y\right)}=0$
This is true if, and only if, $y=0$.

No, since the next-to-last expression isn't even defined if y = 0.

In the next-to-last expression, combine the fractions. Keep in mind that $\ln^2(y^2)$ means $[\ln(y^2)]^2$, and simplify the $\ln(y^2)$ part. You'll see that F'(y) = 0 for a value of y > 0.

 

[Orodruin]

I'm going on the basis of the original problem statement, which says nothing about x and y being independent.

You cannot have an integration boundary that depends on the integration variable. That makes no sense. The relevant equations quoted are correct, he has just not differentiated correctly (although $x$ and $y$ are mixed up in the relevant equations relative to what is in the problem statement). Even if it was the case that $y$ was not in the integration boundary, there is no a priori reason to assume that it depends on $x$. It is stated in the thread title that the final integral should depend on a parameter, clearly $y$. Not that it matters here, the integrand is independent of $y$ anyway.

No, since the next-to-last expression isn't even defined if y = 0.

This is not the point, the point is that the derivative is not taken correctly. If the derivative is taken correctly, he will not have this problem.

For this problem $F'(y) = \frac d{dy} \left( \int_{y}^{y^{2}}\frac{1}{\ln^{2}x}dx\right)$, and my approach would be exactly as I said in post #3.

But, again, your approach is exactly equivalent to the OP's approach. The only difference is that you probably differentiated correctly.

 

[Mark44]

You cannot have an integration boundary that depends on the integration variable.

I don't understand why you're mentioning this -- the problem in post #1 is $F(y) = \int_y^{y^2}\frac 1 {\ln^2(x)}dx$.

The variable of integration (or dummy variable) is x. The limits of integration are y and y². These limits make the integral a function of y.

Problems of this type are standard in 1st year calculus courses, about the time that the Fund. Thm. of Calculus is presented.

 

[Orodruin]

I don't understand why you're mentioning this -- the problem in post #1 is $F(y) = \int_y^{y^2}\frac 1 {\ln^2(x)}dx$.

The variable of integration (or dummy variable) is x. The limits of integration are y and y². These limits make the integral a function of y.

Because you seemed to imply that $y$ should be treated as a function of $x$.

You can assume that y is a function of a single variable: x; i.e., y = f(x).

 

[Mark44]

Because you seemed to imply that $y$ should be treated as a function of $x$.

I added to my previous post, after you replied. The problem here is typical of first-year calculus courses. There is no reason that I can think of that partial derivatives need to be invoked, and I don't see any harm in assuming that y is a function of x, the dummy variable in this integration.

 

[Orodruin]

I added to my previous post, after you replied. The problem here is typical of first-year calculus courses. There is no reason that I can think of that partial derivatives need to be invoked, and I don't see any harm in assuming that y is a function of x, the dummy variable in this integration.

In the problem itself there is no need for partial derivatives. In the relevant equation that the OP wrote down there is. Regardless, you cannot have an integration boundary that depends on the integration variable. To me you seem to be suggesting that the OP write the integral as
$$
\int_{y(x)}^{y(x)^2} \ln^2(x^2) dx,
$$
which clearly makes no sense.

 

[Mark44]

In the problem itself there is no need for partial derivatives. In the relevant equation that the OP wrote down there is. Regardless, you cannot have an integration boundary that depends on the integration variable. To me you seem to be suggesting that the OP write the integral as
$$
\int_{y(x)}^{y(x)^2} \ln^2(x^2) dx,
$$
which clearly makes no sense.

OK. I see what you're saying, so I'll retract my assumption. My objection to the first post was the use of partial derivatives where they are neither called for nor needed, and in particularly in writing the integrand as f(x, y).

 

[Peter Alexander]

I think I managed to solve the task. The result is $y=2$, and this is how I got it.

As @Orodruin has pointed out, the mistake lays in a miscomputation of the lower boundary of integration. It is a superficial mistake I don't know how I managed to overlook.
Because $\frac{d}{dy} y = 1$, the equation, when simplified, turns out being $$\frac{y-2}{\ln{y}}=0$$ which obviously means that $y=2$.

On the final note, I would like to thank both of you, @Orodruin and @Mark44 for taking a look at this problem. It did help me a lot and I appreciate you taking time to help me.