Laplace Transform of power of an integral

Question

We know: laplace transform of $\int_{0}^{t}f(x)\,dx$ with respect to s is $\frac{Laplace\,transform(f(t),\,s)}{s}$

What is laplace transform of $(\int_{0}^{t}f(x)\,dx)^{m}$ ?

Answer 1

I dont think you want to do this: $$ \int_{0}^t f(t)dt \to \frac{F(s)}s\\ f(t)g(t) \to \frac1{i2\pi}\int_{\sigma=-\infty}^\infty F(\sigma)G(s-\sigma)d\sigma\\ f^2(t) \to \frac1{i2\pi}\int_{\sigma=-\infty}^\infty F(\sigma)F(s-\sigma)d\sigma\\ f^n(t) \to \frac1{(i2\pi)^n} \int_{\sigma_2=-\infty}^\infty ...\int_{\sigma_1=-\infty}^\infty F(\sigma_1)F(\sigma_2-\sigma_1)... F(s-\sigma_n)d\sigma_1...d\sigma_n $$ Finally: $$ \left(\int_{0}^t f(t)dt\right)^n \to \frac1{(i2\pi)^n} \int_{\sigma_2=-\infty}^\infty ...\int_{\sigma_1=-\infty}^\infty \frac{F(\sigma_1)}{\sigma_1}\frac{F(\sigma_2-\sigma_1)}{\sigma_2-\sigma_1}... \frac{F(s-\sigma_n)}{s-\sigma_n}d\sigma_1...d\sigma_n $$

The integrals are calculated on the ROC with $\sigma_i$ constant for each step.