Help with the Laplace transform of (te^(6t) with s>6 (such that the integral converges)

[BenF77]

I know the answer is 1/(s-6)^2. I know I am to integrate (e^(-st)*t*e^(6t)*dt) from 0 to infinity which is = integrate (e^((6-s)t)*t*dt) from 0 to infinity. I am certain that I am to use integration by parts with u = t, du = dt, dv = e^((6-s)t)*dt, and v = (this is where I am uncertain) 1/(6-s) *e((6-s)t). When I find v, do I still have to multiply by -1 because s = f(t) (chain rule)? Integrating this thing is not easy. Thank you for the help.

 

[nasi112]

[MATH]F(s) = \int_{0}^{\infty} e^{-st} \cdot te^{6t} \ dt = \int_{0}^{\infty} te^{(6-s)t} \ dt[/MATH]
[MATH]u = t[/MATH]
[MATH]du = dt[/MATH]
[MATH]v = \frac{e^{(6 - s)t}}{6 - s}[/MATH]
[MATH]dv = e^{(6 - s)t} dt[/MATH]
When you want to find [MATH]v[/MATH], you don't need to multiply by [MATH]-1[/MATH].

You need to do this.

[MATH]\int e^{(6 - s)t} \ dt = \int \frac{(6 - s)e^{(6 - s)t}}{(6 - s)} \ dt = \frac{1}{(6 - s)}\int (6 - s)e^{(6 - s)t} \ dt = \frac{1}{6 - s} \cdot e^{(6 - s)t} + C[/MATH]
But of course, you don't need the [MATH]C[/MATH] when you are using integration by parts.