Solving a tricky integral in a simple way

Question

I've been told that this tricky integral is fairly easy to solve with Laplace transform:

However, I am not sure on the solution itself after taking the Laplace transform of the integral. Could someone please help to arrive at the full solution?

Answer 1

Since I is the convolution of f(x) = x^2 and g(x) = cos(3x), you probably figured out that

L[I](s) = L[f](s) L[g](s) = 2/(s^2(s^2+9)). N

ow we want to take the inverse laplace transform, but first we should change this function into a slightly nicer form. The common way to do this is using partial fractions:

we know that 2/(s^2(s^2+9)) = A/s+B/s^2+(Cs+D)/(s^2+9).

It's not to bad to apply the inverse laplace transform to each of these terms, and L^-1 is linear, so once we solve for A,B,C,D, we'll pretty much be done. In the end you should get the same answer as Yefim got above using integration by parts :)

Answer 2

u = (x - y)²; du = - 2(x - y)dy; dv = cos(3y)dy; v = 1/3sin(3y); I = 1/3(x - y)²sin(3y)₀^x + ∫₀^x2/3(x - y)sin(3y)dy =

-2/9(x - y)cos(3y)₀^x ∫₀^x - ∫₀^x2/9cos(3y)dy = 2/9x - 2/27sin(3y)₀^x = 2/9x - 2/27sin(3x)

2 times by parts