Integrate $\int_{0}^{\infty} (\frac{1}{1-jq}) e^{ \frac{j Nq}{1-jq} x} dx$

Question

I want to solve: $\int_{0}^{\infty} (\frac{1}{1-jq}) e^{\frac{jNq}{1-jq} x} dx$

Where $j = \sqrt{-1}$ and $q \in \mathbb{R}$. I would like to see if my approach is correct, as Maple is giving me a different answer.

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The first thing to do is notice that:

$$ \frac{q}{1-jq} = \frac{-q}{jq-1}$$

This gives us the equivalent integral:

$$\int_{0}^{\infty}\Big(\frac{-1}{jq-1}\Big) e^{\frac{-jNq}{jq-1} x} dx$$

We now notice that:

$$\frac{d}{d x} e^{\frac{-jNq}{jq-1} x} = \frac{-jNq}{jq-1} e^{j N \frac{-q}{jq-1} x}$$

So we multiply the integral by $\frac{jNq}{jNq}$, giving us:

$$\frac{1}{jNq} \int_{0}^{\infty}\Big(\frac{-jNq}{jq-1}\Big) e^{ \frac{-j Nq}{jq-1} x} dx$$

Thus if we let $f(x) = e^{ \frac{-j Nq}{jq-1} x}$, we have:

$$\frac{1}{jNq} \int_{0}^{\infty}\Big(\frac{-jNq}{jq-1}\Big) e^{ \frac{-j Nq}{jq-1} x} dx = \frac{1}{jNq}\Big( \lim_{x \rightarrow \infty} f(x) - f(0) \Big)$$ $$\frac{1}{jNq}\Big( \lim_{x \rightarrow \infty} f(x) - f(0) \Big) = \frac{1}{jNq}\Big( \lim_{x \rightarrow \infty} e^{ \frac{-j Nq}{jq-1} x} - 1 \Big)$$

We evaluate the limit now. We see that:

$$ \lim_{x \rightarrow \infty} |e^{ \frac{-j Nq}{jq-1} x}| = \lim_{x \rightarrow \infty} e^{-\frac{Nq^{2}x}{q^{2}+1}} = 0 $$

Therefore:

$$\frac{1}{jNq} \int_{0}^{\infty}\Big(\frac{-jNq}{jq-1}\Big) e^{ \frac{-j Nq}{jq-1} x} dx = \frac{1}{jNq}\Big( 0 - 1 \Big) = \frac{-1}{jNq} = \frac{j}{Nq}$$

So finally:

$$\int_{0}^{\infty} (\frac{1}{1-jq}) e^{\frac{jNq}{1-jq} x} dx = \frac{j}{Nq} $$

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Is this correct? Am I missing something?

Thanks for your time!

Answer 1

The complex exponential makes things a tad bit more complicated. First, we simplify: \begin{align*} \frac{jNq}{1-jq}\,x&=\frac{1+jq}{1+jq}\cdot\frac{jNq}{1-jq}\,x \\ &=\frac{-Nq^2+jNq}{1+q^2}\,x. \end{align*} The real part is $$\frac{-Nq^2}{1+q^2}\,x, $$ so depending on the signs of these variables, the integral may or may not converge. If the coefficient of $x$ is negative, the integral will converge. Assuming that, we have \begin{align*} \int_{0}^{\infty} \left(\frac{1}{1-jq}\right) e^{\frac{jNq}{1-jq} x} \,dx &=\left(\frac{1}{1-jq}\right)\int_{0}^{\infty} e^{\frac{jNq}{1-jq} x} \,dx \\ &=\left(\frac{1}{1-jq}\right)\left(\frac{1-jq}{jNq}\right) e^{\frac{jNq}{1-jq} x}\bigg|_0^{\infty} \\ &=-\left(\frac{j}{Nq}\right) \exp\left(\frac{-Nq^2+jNq}{1+q^2}\,x\right)\bigg|_0^{\infty} \\ &=\frac{j}{Nq}. \end{align*}