But, $$\begin{aligned} &\left|e^{i k x-a x}\right| \\=& R e\left(e^{-a x}(\cos k x-i \sin k x)\right) \\=& e^{-a x} \cos k x \end{aligned}$$

The first equality is false.

can you please elaborate?

Even if you were right, since $\lim_{x\to\infty}e^{-ax}=0$ and $\cos(kx)$ is bounded, we would still have $\lim_{x\to\infty}e^{-ax}\cos(kx)=0$.

You seem to think that $|e^z|=\operatorname{Re}(e^z)$. This is not true.

then, @José Carlos Santos, what does the | | operator do?

That is the absolute value.

That is $|x+yi|=\sqrt{x^2+y^2}$.

But Sir, there is no absolute value is used in the solution. Then why you took the absolute value?

Because proving that $\lim_{x\to\infty}f(x)=0$ is the same thing as proving that $\lim_{x\to\infty}|f(x)|=0$. But proving that the second equality holds is easier, since it deals only with non-negative real numbers.

Sir, you have already considered that, $lim_{x \rightarrow \infty} e^{(ik-a)x}=0$, so you have took the absolute value to proove again that $\lim_{x\to\infty}|f(x)|=0. But if we completely don't know the value of $lim_{x \rightarrow \infty} e^{(ik-a)x}$, then how to step ahead in the solution?

I don't know we you claim that I already know that $\lim_{x\to\infty}e^{ikx-ax}=0$. I deduced that from the fact that $\lim_{x\to\infty}|e^{ikx-ax}|=0$.

It seems to me that you don't like my answer. Do you want me to delete it?

I am saying that, if we don't know the result, then there is no benefit of taking the absolute value (to prove it is 0) as the result could be any unknown value . So, can it be proved to be 0 without taking the absolute value?

Sure. Since $e^{ikx-ax}=e^{-ax}\cos(kx)+e^{-ax}\sin(kx)i$, just use the fact that $\lim_{x\to\infty}e^{-ax}=0$, together with the fact that both functions $\cos(kx)$ and $\sin(kx)$ are bounded. But you are wrong when you claim that “there is no benefit of taking the absolute value”. It makes things easier.

Can it be shown that, $\left[\ln \left(1-e^{-x}\right)\right]_{0}^{\infty}=0$ ?

Actually I faced this problem during work (which, if true, leads to a correct consequence). I found the similar problem in the book thich seemed to be infinite to me at first. So I asked that question there.

No. Actually, $\left[\ln\left(1-e^{-x}\right)\right]_0^{\infty}=\infty$. And please don't ask more questions which are not related to my answer. Again: do you want me to delete it?

ok sir. Thank You