I'm trying to calculate this integral, and I find it difficult when coping with complex numbers.
$$ f(k) = \int_{lnK}^{\infty} e^{ikx} (e^{x}-K) dx =(\frac{e^{(ik+1)x}}{ik+1}-K\frac{e^{ikx}}{ik})|_{x=lnK}^{x=\infty} $$
k is a complex number, Im k>1
and this equation should be equal to $$ -\frac{K^{1+ik}}{k^2-ik} $$ which means the result of the upper limit is 0.
What I cannot understand is that when an infinity multiplied by a complex number, is it still an infinity? It seems like a weird question, and I'm not sure whether I make myself understood. I mean, since exp(-infinity)=0, is exp(a complex number *-infinity) still equals to zero?
Many thanks!
If $k=\alpha+i\beta$ with $\beta>1$ then $$\left|e^{(ik+1)x}\right|=\left|e^{i\alpha x}\>e^{(1-\beta) x}\right|= \bigl|e^{(1-\beta) x}\bigr|\to0\qquad(x\to\infty)\ .$$ It follows that the improper integral $$\int_a^\infty e^{(ik+1)x}\>dx:=\lim_{b\to\infty}\int_a^b e^{(ik+1)x}\>dx\tag{1}$$ is convergent, and there is no "multiplication with $\infty$" happening. Just compute the $\int_a^b$ on the RHS of $(1)$ the ordinary way, and you will see that the $\lim_{b\to\infty}$ of this integral exists.