Convergence of $\int\limits^0_{-\infty} e^{i a t + b t} dt$

Question

Consider the integral $I :=$ $\int^{0}_{-t_{0}} e^{iat+bt} dt$, then it is true that $$ I = \frac{(1-e^{-(ia+b)t_{0}})}{ia+b} $$

I am now interested on the case where $ {t_{0}\to -\infty} $. Wolfram Alpha says that for $Re(b)>Im(a)$ it converges to $I = \frac{1}{ia+b} $

For my case I even have $a,b \in \mathbb{R}$ but I am still interested why this is the case. A hint where or at what to look would be very helpful, thanks!

Answer 1

Lets just consider $z\in\mathbb C$ non-zero and write $$I(t_0)=\int_{-t_0}^0e^{zt}dt$$ for $t_0\geq0$. Then, as you noted, $$I(t_0)=\frac{1-e^{-zt_0}}{z}.$$ Now we have $$\left|\frac{1}{z}-\frac{1-e^{-zt_0}}{z}\right|=\frac{|e^{-zt_0}|}{|z|}=\frac{e^{-Re(z)t_0}}{|z|}.$$ So if $Re(z)>0$ the right-hand side goes to zero, hence $I(t_0)$ converges to $\frac{1}{z}$ as $t_0\to\infty$.
Now if we write $z=ia+b$ where $a,b\in \mathbb C$, then $Re(z)=Re(b)-Im(a)$, so the condition $Re(z)>0$ is equivalent to $Re(b)>Im(a)$.

Answer 2

If $a,b \in \mathbb{R}$.

$$|e^{(a+bi)t}|=e^{at}|e^{bi t}|=e^{at}$$

So long as $a>0$, then the expression tends to $0+0i$ as $t \to -\infty$ no matter the value of $b$. Otherwise if $a=0$ then the expression fluctuates as sine and cosine fluctuate. If $a<0$ then because the magnitude of the complex number $e^{(a+bi)t}$ diverges, then so does the complex number itself.

So if $\Re z>0$ then $e^{zt}$ converges, otherwise it diverges.

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If $z=b+ai=\Re b+i \Im b+i \Re a-\Im a$

Then $\Re z>0$ if and only if $\Re b> \Im a$.