# Laplace Transform Shifting Property

I am not quite sure if I understand the frequency shifting property of the Laplace Transform.

If I have

$$I_1(a,s)=\int_0^{\infty}\mathrm{e}^{-a\,t}f(t)\,\mathrm{e}^{-s\,t}\, \mathrm{d}t$$

And I do know

$$I_2(s)=\int_0^{\infty}f(t) \mathrm{e}^{-s\,t}\, \mathrm{d}t$$

Is $$I_1(a,s)=I_2(s+a)$$ independend if $$a$$ is positiv or negativ?

If this is the case, is it possible to solve

$$I_4(\gamma,b,\beta)=\pi \,\sqrt{\beta^2+b^2}\int_{0}^{\infty} e^{-\omega \frac{\gamma \, \beta}{\sqrt{b^2+\beta^2}}} \, \frac{1}{\sqrt{\omega^2+\beta^2+b^2}} \, \mathrm{K}_1\left(\gamma \, \sqrt{\omega^2+\beta^2+b^2}\right) \mathrm{d}\omega$$ if I do know $$I_3(\gamma,b,\beta)=\pi \,\sqrt{\beta^2+b^2}\int_{0}^{\infty} e^{-\omega \gamma } \, \frac{1}{\sqrt{\omega^2+\beta^2+b^2}} \, \mathrm{K}_1\left(\gamma \, \sqrt{\omega^2+\beta^2+b^2}\right) \mathrm{d}\omega$$

The original problem was shown here.

As far as I understand it, the answer to your first question is yes:

$$I_1(a,s)=\int_0^{\infty}e^{-at}f(t)e^{-st}\, dt$$

$$I_1(a,s)=\int_0^{\infty}e^{-(s+a)t}f(t)\, dt$$

Since

$$I_2(x)=\int_0^{\infty}e^{-xt}f(t)\, dt$$

$$I_2(s+a)=\int_0^{\infty}e^{-(s+a)t}f(t)\, dt$$

which is $I_1(a,s)$. Note that we never used $a<0$ or $a>0$, so as long as the integral $I_1(a,s)$ converges, we're fine.

With regards to your second question, that's a little more tricky, since the function that would be $f(\omega)$ depends on the parameter $\gamma$, which is the one we'd like to shift. Note that the Laplace Transform shifting property you mentioned is only valid when $f$ is independent of $s$.