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For any function f continuous on $\,(-\infty\,,\,\infty)\,$:

$$g(x) = \int_0^x f(t)\,dt$$

$$h(x) = \int_0^x (x-t)f(t)\,dt$$

$$w(x) = \int_0^x f(t)\sin(x-t)\,dt$$

Show that

$$h(x) = \int_0^x g(u)\, du$$


$$ \frac{d^2w}{dx^2} + w = f(x) w(0)=0\,\,,\,\text{and}\,\, w'(0) = 0$$

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First, show us some self work, ideas, effort....second, it'd be a good idea if you go to the FAQ section and learn there a little about how to properly write mathematics with LaTeX in this site. – DonAntonio Dec 3 '12 at 12:47
Also please use a meaningful title so that others can find your question if they are looking for something similar themselves. – Simon Hayward Dec 3 '12 at 13:02
up vote 4 down vote accepted

For the first, note that

$$h(x) = \int_0^x (x-t)f(t)\,dt$$

$$h(x) =x \int_0^xf(t)\,dt- \int_0^x tf(t)\,dt$$

so that

$$h'(x) = \int_0^xf(t)\,dt+ x f(x)- x f(x)= \int_0^xf(t)\,dt=g(x)$$

We used FTC and the product rule.


$$h(x)-h(0)=h(x)=\int_0^x g(u)du$$

For the second one we need a little trickery

$$w(x) = \int_0^x f(t)\sin(x-t)\,dt$$

$$w(x) = \int_0^x f(t)(\sin x \cos t -\sin x \cos t)\,dt$$

$$w(x) =\sin x \int_0^x f(t) \cos t \,dt-\cos x \int_0^x f(t)\sin t\,dt$$

Now differentiate, using the product rule and FTC.

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And sorry for the improper writing style for maths. Actually, I have no idea in this type of questions and it confused me a lot. THXx10000 – hugo Dec 3 '12 at 12:58

There is actually no trickery needed : you can differentiate $w(x)$ using the chain rule. If it helps, note $F(x,y) = \int_0^y f(t)\sin (x-t)\,dt$ and remark that $w(x)=F(x,x)$. For example, one obtains $$ w'(x) = \int_0^x f(t) \cos (x-t)\,dt + f(x)\sin(x-x) = \int_0^x f(t)\cos (x-t). $$ Now, you should be able to compute $w''(x)$.

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I finally computed w''(x)=f(x) Is it wrong so I cannote prove d^2w/dx^2 +w = f(x)? – hugo Dec 3 '12 at 13:23
Indeed,$w''(x) = - \int_0^x f(t)\sin(x-t)\,dt + f(x) \cos (x-x) = f(x) - w(x)$. – Siméon Dec 3 '12 at 13:38
I don't understand why there is a "$$\int_0^x -f(t)sin(x-t)\,dt$$" – hugo Dec 3 '12 at 13:43
I got it, but I still miss -. I computed w(x)+f(x) ~.~ But I think it may be my careless mistakes~thanks – hugo Dec 3 '12 at 13:54
What is $\frac{d}{dx} \cos(x)$ ? – Siméon Dec 3 '12 at 14:02

Here is the solution for the first part $$ h(x) = \int_0^x g(u)du = \int_{0}^{x} \int_{0}^{u}f(t) \, dt du = \int_{0}^{x} \int_{t}^{x}f(t) \, du dt = \int_{0}^{x} (x-t)f(t) dt.$$

In the above, we changed the order of integration. To see that, plot the region $0<u<x, \,0<t<u$.

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How do you get to $$\int_0^x \int_t^x$$ in the third step? – Pedro Tamaroff Dec 3 '12 at 13:01
@PeterTamaroff: I changed the variable of integration to $t$. – Mhenni Benghorbal Dec 3 '12 at 13:07
Could you clarify in the post? I don't get how you get those new limits. – Pedro Tamaroff Dec 3 '12 at 14:47

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