Improper integrals with trigonometric function

Question

$$P.V.\int_{-\infty}^\infty \frac{\cos 3x}{x^2+4}dx=\operatorname{Re} P.V.\int_{-\infty}^\infty \frac{e^{i3x}}{x^2+4}dx$$

P.V.=(principal value)

I need help understanding why this "trick" works and when it is applicable.

I know that: $$P.V.\int_{-\infty}^\infty \frac{\cos 3x}{x^2+4}dx=P.V.\frac{1}{2}\int_{-\infty}^\infty \frac{e^{i3x}}{x^2+4}dx+P.V.\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-i3x}}{x^2+4}dx$$

Answer 1

You just use the euler identity $$ e^{zi}=\cos(z)+\sin(z)i. $$ If $z\in\mathbb R$, you get $\sin(z),\cos(z)\in\mathbb R$ and $Re(e^{zi})=\cos(z)$.

Answer 2

This "trick" works because $$ \cos x= \text{Re}\; e^{ix} $$ for all $x\in\mathbb{R}$.

Answer 3

This follows from Euler's Formula

$$e^{ix}=\cos{x}+i\sin{x}$$ $Re$, sometimes written as $\Re$ is just the real part of a complex number (which in the case of $e^{ix}$ is $\cos{x}$). By substitution:

$$\frac{\cos{3x}}{x^2+4}=\Re\left(\frac{e^{i3x}}{x^2+4}\right)$$

The reason why the second line works is that for any complex number $z$, the real part may be obtained by:

$$\Re(z)=\frac{z+\overline{z}}{2}$$

Where $\overline{z}$ is the complex conjugate. As $e^{i3x}$ is the complex conjugate of $e^{-i3x}$, their sum, divided by two, is equal to the real part of either one.

Answer 4

Understand your question did not ask for this specifically but I figure it can never hurt to have other methods.

Here you wish to solve the integral:

\begin{equation} I = \int_{-\infty}^{\infty} \frac{\cos(3x)}{x^2 + 4}\:dx \end{equation}

Let's instead consider the generalised case: \begin{equation} J(y,z) = \int_{-\infty}^{\infty} \frac{\cos(yx)}{x^2 + z^2}\:dx \end{equation}

We observe that $I = J(3,2)$.

First we observe that the integrand's parity is even, thus

\begin{equation} J(y,z) = 2\int_{0}^{\infty} \frac{\cos(yx)}{x^2 + z^2}\:dx \end{equation}

We now Employ Feyman's Trick with Laplace Transforms. We first introduce a parameter $t$ such that:

\begin{equation} H(t; y,z) = 2\int_{0}^{\infty} \frac{\cos(yxt)}{x^2 + z^2}\:dx \end{equation}

We observe that $J(y,z) = \lim_{t \rightarrow 1+}$. We now observe that the integral complies with the Dominated Convergence Theorem and Fubini's Theorem. We now take the Laplace Transform with respect to $t$:

\begin{align} \mathscr{L}_t\left[H(t; y,z) \right] &= 2\int_{0}^{\infty} \frac{\mathscr{L}_t\left[\cos(yxt)\right]}{x^2 + z^2}\:dx = 2\int_{0}^{\infty} \frac{s}{s^2 + x^2} \frac{1}{x^2 + z^2}\:dx = 2\int_{0}^{\infty} \frac{s}{\left(s^2 + y^2x^2\right)\left(x^2 + z^2\right)}\:dx \\ &= \frac{2s}{s^2 - y^2z^2}\left[\int_{0}^{\infty} \frac{-y^2}{s^2 + y^2x^2}\:dx + \int_{0}^{\infty} \frac{1}{z^2 + x^2}\:dx \right] \\ &= \frac{2s}{s^2 - y^2z^2}\left[ -y^2\frac{1}{\left| sy\right|}\arctan\left(\left|\frac{y}{s}\right|x \right)\bigg|_{0}^{\infty} + \frac{1}{|z|}\arctan\left(\frac{x}{|z|} \right)\bigg|_{0}^{\infty}\right] =\frac{2s}{s^2 - y^2z^2}\left[ -y^2\frac{1}{\left| sy\right|}\frac{\pi}{2} + \frac{1}{|z|}\frac{\pi}{2}\right]\\ &= \frac{\pi s}{s^2 - y^2z^2} \frac{|s| - |zy|}{|sz|} = \frac{\pi}{|z|\left(s +|yz| \right)} \end{align}

we now take the inverse Laplace Transform:

\begin{equation} H(t; y,z) = \mathscr{L}_t^{-1}\left[ \frac{\pi}{|z|\left(s +|yz| \right)} \right] = \frac{\pi}{|z|}e^{-|yz|t} \end{equation}

Hence we now can solve $J(y,z)$

\begin{equation} J(y,z) = \lim_{t \rightarrow 1+} H(t;y,z)= \lim_{t \rightarrow 1+}\frac{\pi}{|z|}e^{-|yz|t} = \frac{\pi}{|z|}e^{-|yz|} \end{equation}

Finally:

\begin{equation} I = J(3,2) = \frac{\pi}{|2|}e^{-|2\cdot3|} = -\frac{\pi}{2}e^{-6} \end{equation}