Derivative of Integral: How to Use the Fundamental Theorem of Calculus

[dumbQuestion]

Homework Statement

Find d/dx[∫⁰_x(cos(xt)/t)dt] (just in case the notation is not clear, the differential with respect to x, of the definite integral from x to 0, of cos(xt) over t)

Homework Equations

My initial thought was just using the fundamental theorem of calculus

The Attempt at a Solution

Let I(t)=∫⁰_x(cos(xt)/t)dt

Now, let f(t)=cos(xt)/t (the integrand) and let F(t) be an antiderivative of f(t) (a thought: f(t) is not defined at t=0, is this where I'm going wrong? Can I define F(t) to be an antiderivative? I mean, f(t) is not defined at t=0 but it's still integrable everywhere right?). Then by the fundamental theorem, I(t)=F(0)-F(x)

So now we can take d/dx of both sides

d/dx(I(t)) = d/dx[F(0) - F(x)]
d/dx[∫⁰_x(cos(xt)/t)dt] = f(0) - f(x) (by FTC)
d/dx[∫⁰_x(cos(xt)/t)dt] = cos(0)/0 - cos(x²)/x

In this step I can't proceed because I'm dividing by zero. If I try to take the limit f(t) as t --> 0 the denominator blows up and the whole thing goes to infinity, so this limit does not exist. So it seems like this is an improper integral? I'm confused.~~~~~~~~~~~~~```
However, the book gives as the answer (1/x)(1-2cos(x²)) and I don't believe they use the FTC. I can use their method to just find the integral but I would like to see how to use the FTC in this case

 

[Mark44]

Homework Statement

Find d/dx[∫⁰_x(cos(xt)/t)dt] (just in case the notation is not clear, the differential with respect to x, of the definite integral from x to 0, of cos(xt) over t)

Here's the LaTeX version of the integral. You can click what I wrote to see how I did it.

$$ \int_{t = x}^0 \frac{cos(xt) dt}{t}$$

Homework Equations

My initial thought was just using the fundamental theorem of calculus

The Attempt at a Solution

Let I(t)=∫⁰_x(cos(xt)/t)dt

Now, let f(t)=cos(xt)/t (the integrand) and let F(t) be an antiderivative of f(t) (a thought: f(t) is not defined at t=0, is this where I'm going wrong? Can I define F(t) to be an antiderivative? I mean, f(t) is not defined at t=0 but it's still integrable everywhere right?). Then by the fundamental theorem, I(t)=F(0)-F(x)

So now we can take d/dx of both sides

d/dx(I(t)) = d/dx[F(0) - F(x)]
d/dx[∫⁰_x(cos(xt)/t)dt] = f(0) - f(x) (by FTC)
d/dx[∫⁰_x(cos(xt)/t)dt] = cos(0)/0 - cos(x²)/x

In this step I can't proceed because I'm dividing by zero. If I try to take the limit f(t) as t --> 0 the denominator blows up and the whole thing goes to infinity, so this limit does not exist. So it seems like this is an improper integral? I'm confused.


~~~~~~~~~~~~~```
However, the book gives as the answer (1/x)(1-2cos(x²)) and I don't believe they use the FTC. I can use their method to just find the integral but I would like to see how to use the FTC in this case

I would work with this integral, and take the limit of it as a approaches 0.
$$ \int_{t = x}^a \frac{cos(xt) dt}{t}$$

I haven't worked this through, but this is what I would try first.

 

[STEMucator]

Just as mark said up there ^ I believe you'll be able to use the squeeze theorem afterwards to clean up you problem.

 

[dumbQuestion]

Thanks guys. I am going to try this approach. Just out of curiosity, is there a reason I shouldn't apply the FTC in this problem? I mean, it appears its easier to do it this way that has been suggested to me in the responses, but is it still feasible to apply the FTC or is there a reason why that's not allowed in this case? I guess I'm asking, can I use the FTC when the integrand has a singularity at one of the end points of the interval I"m integrating over, or within the interval itself?

 

[STEMucator]

The FTC allows you to evaluate the integral at the endpoints. The endpoints themselves are the cause of the problem, not the process in which you got to evaluating the endpoints. So yes you can still use the FTC, just pull the endpoint out as a limit so you can evaluate the integral properly and then check the limit after you're done.

 

[hedipaldi]

sorry for using attachment.I am working on my latex,hoping to be able to use it soon.

 

[Dick]

Thanks guys. I am going to try this approach. Just out of curiosity, is there a reason I shouldn't apply the FTC in this problem? I mean, it appears its easier to do it this way that has been suggested to me in the responses, but is it still feasible to apply the FTC or is there a reason why that's not allowed in this case? I guess I'm asking, can I use the FTC when the integrand has a singularity at one of the end points of the interval I"m integrating over, or within the interval itself?

The complication that keeps you from simply using FTC is that you have x dependence in the integrand. You need to use http://en.wikipedia.org/wiki/Leibniz_integral_rule
That will give you the book answer. That said, I think the exercise is a little silly since the integral does not exist for any nonzero value of x.

 

[hedipaldi]

for x and a greater than a positive number,there is no singularity in the interval of integration

 

[Dick]

for x and a greater than a positive number,there is no singularity in the interval of integration

That's if you do a cutoff at a then let a->0. That let's you get a derivative that exists but the original integral (the thing that you are taking the derivative of) still doesn't exist. You do this sort of thing in physics, but it seems a little odd here.

 

[arildno]

FTC is precisely what you use here.
It's just that it uglifies, into Leibniz differentiation, due to its "essential" two-variable nature.

 

[hedipaldi]

this integral does converge,by Dirichle criterion.Examine uniform convergence for using Leibnitz formula.

 

[Dick]

this integral does converge,by Dirichle criterion.Examine uniform convergence for using Leibnitz formula.

I'm not sure I believe that. Put x=1. The resulting integral doesn't look convergent me by a simple comparison test.

 

[HallsofIvy]

The general Leibnitz formula says that
\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x, t)dt= \frac{d\beta(x)}{dx}f(x,\beta(x))- \frac{d\alpha(x)}{dx}f(x,\alpha(x))+ \int_{\alpha(x)}^{\beta(x)}\frac{\partial f(x,t)}{\partial x}dx

 

[arildno]

It is, most definitely, convergent

 

[dumbQuestion]

Thank you guys I am finding this discussing very useful and hedipaldi, thank you so much for taking the time to upload that document. This is all so appreciated by me!

 

[Dick]

It is, most definitely, convergent

The formula for the derivative is convergent, if that's what you mean. The integral you are taking the derivative of isn't. Most definitely.

 

[hedipaldi]

yes,i agree.

 

[hedipaldi]

the initial integral converges al the infinity but not at 0,so the differentiation formula is valid only when the interval does not contain 0