Hello I am struggling with the following integral:
$$ I(t) = \displaystyle\int\limits_{0}^{\infty} e^{jx(t+j)}dx $$
Where $j = \sqrt{-1}$. I think the answer should be:
$$ I(t) = \frac{1}{1-jt} $$
However I don't understand how to evaluate the limit in the following:
$$ I(t) = \frac{1}{j(t+j)} \; \big( \underset{x \rightarrow \infty}{\lim} e^{jx(t+j)} -1 \big)$$
I tried expanding it using the Euler formula, but it doesn't help because the cosine doesn't have a limit, right? Any help?
On
Well, in a more general sense we're trying to solve:
$$\mathcal{I}_\text{n}:=\lim_{\eta\to\infty}\int_0^\eta\exp\left(\text{n}\cdot x\right)\space\text{d}x\tag1$$
Substitute $\text{u}=\text{n}\cdot x$, so we get:
$$\mathcal{I}_\text{n}=\frac{1}{\text{n}}\cdot\lim_{\eta\to\infty}\int_0^{\text{n}\eta}\exp\left(\text{u}\right)\space\text{d}\text{u}=\frac{1}{\text{n}}\cdot\lim_{\eta\to\infty}\left[\exp\left(\text{u}\right)\right]_0^{\text{n}\eta}=$$ $$\frac{1}{\text{n}}\cdot\lim_{\eta\to\infty}\left(\exp\left(\text{n}\cdot\eta\right)-\exp\left(0\right)\right)=\frac{1}{\text{n}}\cdot\lim_{\eta\to\infty}\left(\exp\left(\text{n}\cdot\eta\right)-1\right)\tag2$$
So, when $\text{n}=i(t+i)$ we get (when $t\in\mathbb{R}$):
$$\mathcal{I}_{i(t+i)}=\frac{1}{i(t+i)}\cdot\lim_{\eta\to\infty}\left(\exp\left(i(t+i)\cdot\eta\right)-1\right)=\frac{1}{i(t+i)}\cdot\left(0-1\right)=\frac{i}{i+t}\tag3$$
Note that $e^{jx(t+j)}=e^{jxt}e^{-x}$, so $|e^{jx(t+j)}|=|e^{-x}|$, hence $$\lim_{x\to\infty}e^{jx(t+j)}=0$$