The Convergence of Complex Integrals

[Ted123]

I know that for any a>0 and k,t\in\mathbb{R}, the integral \int_0^a t^k\; dt converges if and only if k>-1.

Is it true that if k is complex then \displaystyle \int_0^a |t^k| \; dt converges if and only if \text{Re}(k)>-1 since if t is real, |t^k| does not depend on the imaginary part of k?

 

[Dick]

I know that for any a>0 and k,t\in\mathbb{R}, the integral \int_0^a t^k\; dt converges if and only if k>-1.

Is it true that if k is complex then \displaystyle \int_0^a |t^k| \; dt converges if and only if \text{Re}(k)>-1 since if t is real, |t^k| does not depend on the imaginary part of k?

Yes, it is. t^k=e^{log(t) k}. If k is pure imaginary and t>0 |t^k|=1.

 

[Ted123]

Yes, it is. t^k=e^{log(t) k}. If k is pure imaginary and t>0 |t^k|=1.

So would you prove it as follows:

If x+iy=k\in\mathbb{C} and t,x,y\in\mathbb{R} with t>0 then |t^k| = |t^{x+iy}| = |t^x t^{iy}| = |t^x e^{\ln(t)iy}| = |t^x||e^{\ln(t)iy}| = |t^x| = t^x so we're just back in the real case where we know \int_0^a t^x dt converges for x=\text{Re}(k)>-1.

It is true that |t^x| = t^x for any real x and positive real t isn't it?

 

[Dick]

So would you prove it as follows:

If x+iy=k\in\mathbb{C} and t,x,y\in\mathbb{R} with t>0 then |t^k| = |t^{x+iy}| = |t^x t^{iy}| = |t^x e^{\ln(t)iy}| = |t^x||e^{\ln(t)iy}| = |t^x| = t^x so we're just back in the real case where we know \int_0^a t^x dt converges for x=\text{Re}(k)>-1.

It is true that |t^x| = t^x for any real x and positive real t isn't it?

Sure. Why would you doubt that?