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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

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

and

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

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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.2012-12-03
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    Also please use a meaningful title so that others can find your question if they are looking for something similar themselves.2012-12-03

3 Answers 3

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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.

Thus

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

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. THXx100002012-12-03
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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)?2012-12-03
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    Indeed,$w''(x) = - \int_0^x f(t)\sin(x-t)\,dt + f(x) \cos (x-x) = f(x) - w(x)$.2012-12-03
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    I don't understand why there is a "$$\int_0^x -f(t)sin(x-t)\,dt$$"2012-12-03
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    I got it, but I still miss -. I computed w(x)+f(x) ~.~ But I think it may be my careless mistakes~thanks2012-12-03
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    What is $\frac{d}{dx} \cos(x)$ ?2012-12-03
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    I got it!!!!!!thxxxxxxxxxxx T_T I got the wrong answer because I made a mistake in the cos(A-B)angle formula2012-12-03
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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.

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    How do you get to $$\int_0^x \int_t^x$$ in the third step?2012-12-03
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    @PeterTamaroff: I changed the variable of integration to $t$.2012-12-03
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    Could you clarify in the post? I don't get how you get those new limits.2012-12-03