# Thread: differentiation under the integral sign

1. ## differentiation under the integral sign

I have this problem:

With the aid of differentiation under the integral sign, show that the function

y=integral from 0 to x of h(t)sin(x-t)dt

satisfies the differential equation

y''+y=h(t).

Ok. I have the following equation to help me:
the derivative of the integral from a(x) to b(x) of f(x,t)dt is equal to
the integral from a(x) to b(x) of the partial derivative of x in f(x,t)dt + f(x,b(x))*b'(x)-f(x,a(x))*a'(x).

So I note that
a(x)=0
b(x)=x
a'(x)=0
b'(x)=1
f(x,t)=h(t)sin(x-t)
the partial derivative of x of f(x,t)=h(t)cos(x-t)
f(x,b(x))=0
f(x,a(x))=h(0)sin(x)

Thus, the derivative of the integral from a(x) to b(x) of f(x,t)dt is equal to
the integral from 0 to x of h(t)cos(x-t)dt.

This is where I get stuck. I'm pretty sure that if you plug in 0 and x into the above you get the second derivative, but I'm not positive.

Any help would be appreciated.

2. Originally Posted by James234
I have this problem:

With the aid of differentiation under the integral sign, show that the function

y=integral from 0 to x of h(t)sin(x-t)dt
$\displaystyle y=\int_0^x h(t)\sin(x-t)dt$

$\displaystyle y=\int_0^x h(t)(\sin x\cos t-\sin t\cos x ) dt$
$\displaystyle y=\int_0^x h(t)\sin x\cos t-\int_0^x h(t)\cos x\sin t dt$
$\displaystyle y=\sin x\int_0^x h(t)\cos t dt-\cos x \int_0^x h(t)\sin t dt$
$\displaystyle y'=\cos x\int_0^x h(t)\cos tdt +$$\displaystyle h(x)\sin x \cos x+\sin x \int_0^x h(t)\sin t dt-h(x)\sin x \cos x$
$\displaystyle y'=\cos x\int_0^x h(t)\cos tdt+\sin x\int_0^x h(t)\sin tdt$
$\displaystyle y''=-\sin x\int_0^x h(t)\cos tdt+h(x)\cos^2 x+\cos x\int_0^x h(t)\sin tdt+h(x)\sin^2 x$
$\displaystyle y''+y=h(x)\cos^2 x+h(x)\sin^2 x=h(x)(\cos^2 x+\sin^2 x)=h(x)(1)=h(x)$

3. Thank you for the response. Makes a lot of sense.

Do you know the initial conditions that this function has to follow to justify this differentiation?