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?