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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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    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. THXx1$0$0002012-12-03
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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 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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    Could you clarify in the post? I don't get how you get those new limits.2012-12-03