Definite integration and stuff

[Raghav Gupta]

No the leibniz integral rule is:-
$$\frac{\partial}{\partial x}\int_{a(x)}^{b(x)}f(x,t)\ dt=\int_{a(x)}^{b(x)}\frac{\partial f}{\partial x}\ dt+f(b(x),x)\frac{\partial b}{\partial x}-f(a(x),x)\frac{\partial a}{\partial x}$$

 

[Ray Vickson]

What are you talking about ?

I think he is worried about the effects of having $a(x)$ in the lower limit.

 

[Ray Vickson]

Isn't Leibniz rule
$$ \frac{d}{dx} \int_{g(x)}^{h(x)} f(t) = f(h(x))h'(x) - f(g(x)) g'(x) $$ ?
So it should be
$$ \frac{d}{dx} \int_0^x e^{-t} f(x-t) = e^{-x}f(0) ? $$ as second term would be zero because differentiation of a constant which is the lower limit is zero

No. It should be exactly what I wrote in #22.

 

[Raghav Gupta]

$$-f(a(x),x)\frac{\partial a}{\partial x}$$
That last term is not visible in post 25 , that is what I am talking about.
I had to quote to see.

 

[certainly]

It's visible here, @Ray Vickson is it not visible to you too...

 

[Raghav Gupta]

It's visible here, @Ray Vickson is it not visible to you too...

I think I am using a mobile device and you desktop.

 

[Ray Vickson]

It's visible here, @Ray Vickson is it not visible to you too...

Yes, it was visible to me as soon as you posted your response.

 

[Ray Vickson]

I think I am using a mobile device and you desktop.

I have found that on a mobile device, a long LaTeX expression might not display properly. Some of it can run off the right side of the screen, and the device does not start a new line for it. It will start a new line for pure text, but not on a typeset LaTeX/TeX expression.

 

[PeroK]

Isn't Leibniz rule
$$ \frac{d}{dx} \int_{g(x)}^{h(x)} f(t) = f(h(x))h'(x) - f(g(x)) g'(x) $$ ?
So it should be
$$ \frac{d}{dx} \int_0^x e^{-t} f(x-t) = e^{-x}f(0) ? $$ as second term would be zero because differentiation of a constant which is the lower limit is zero

Your fundamental mistake is to treat x as a constant when you are differentiating by x. The definite integral will depend on x not just as a limit of the integral but also in the value of the integrand.

 

[Raghav Gupta]

No the leibniz integral rule is:-
$$\frac{\partial}{\partial x}\int_{a(x)}^{b(x)}f(x,t)\ dt=\int_{a(x)}^{b(x)}\frac{\partial f}{\partial x}\ dt+f(b(x),x)\frac{\partial b}{\partial x}-f(a(x),x)\frac{\partial a}{\partial x}$$

Problems arising for me.
Getting first term in R.H.S
$$\int_0^x e^{-t} f'(x-t)$$ second term 0 and third term also 0 as δa/δx is 0.

 

[certainly]

Problems arising for me.
Getting first term in R.H.S
∫x0e−tf′(x−t)
\int_0^x e^{-t} f'(x-t) second term 0 and third term also 0 as δa/δx is 0.

i'm going to call the function to be integrated $g(x,t)$ (i.e $f$ in the equation you quoted) because it might cause some confusion.
Now, $g(x,t)=e^{-t}f(x-t)$ (this is the $f$ in the original problem)
therefore $g(x,x)\frac{\partial x}{\partial x} = e^{-x}f(0)$
and the third term vanishes not only because $\frac{\partial a}{\partial x} = 0$ but also if look closely you'll notice that the limits of integration are 0 and 0 (EDIT:- In the recursive definition of f), thus making the integral 0 as well. This is a necessary step since the integral could diverge in which case we would have to take the limit.

 

[mooncrater]

Attached

 

[certainly]

It is incorrect to assume that $f(0)=0$...
And if it was mentioned in the question, you should have told us so...

 

[mooncrater]

It is incorrect to assume that $f(0)=0$...
And if it was mentioned in the question, you should have told us so...

No.. it wasn't mentioned in the question.

 

[Raghav Gupta]

Homework Statement

The question says:
f (x)=x²+7x +∫₀^x(e^-tf (x-t)dt.
Find f (x).

But definitely f(0) = 0
As integral from 0 to 0 is zero.

 

[certainly]

errrrr... yes, sorry for that, looks like I'm finally getting sleepy...

 

[Raghav Gupta]

errrrr... yes, sorry for that, looks like I'm finally getting sleepy...

For you .