Help with integration of $\mathrm{exp}(jxt)/(x^2 + a^2)$

Question

I am facing issue in solving this integral: $$ \int_{-\infty}^{\infty} \frac{e^{(jxt)}}{x^2 + a^2} dx $$ I have tried to apply euler formula to seperate $e^{jxt} = cos(xt) + jsin(xt)$ then wolfram gives me result but I don't know how ?

Answer 1

I thought that it would be instructive to present a solution that relies on real analysis only. In the following development, we assume that $a\in \mathbb{R}$ and $t\in \mathbb{R}$.

Let $f(t,a)$ be represented by the integral

$$\begin{align} f(t,a)&=\int_{-\infty}^\infty \frac{e^{ixt}}{x^2+a^2}\,dx \\\\ &=\int_{-\infty}^\infty \frac{\cos(xt)}{x^2+a^2}\,dx+i\int_{-\infty}^\infty \frac{\sin(xt)}{x^2+a^2}\,dx \tag 1 \end{align}$$

Exploiting the even symmetry and odd symmetry of the integrands in the integrals on the right-hand side of $(1)$, we can write

$$f(t,a)=2\int_0^\infty \frac{\cos(xt)}{x^2+a^2}\,dx \tag 2$$

Enforcing the substitution $x\to |a|x$ in $(2)$ reveals

$$\bbox[5px,border:2px solid #C0A000]{f(t,a)=\frac{2}{|a|}\int_0^\infty \frac{\cos(|a|tx)}{x^2+1}\,dx} \tag 3$$

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Next, we use Feynman's Trick for differentiating under the integral and differentiate $f(t,a)$, as given in $(3)$, with respect to $t$. Proceeding, we find that

$$\begin{align} \frac{\partial f(t,a)}{\partial t}&=-2\int_0^\infty \frac{x\sin(|a|tx)}{x^2+1}\,dx\\\\ &=-2\int_0^\infty \frac{(x^2+1)\sin(|a|tx)-\sin(|a|tx)}{x(x^2+1)}\,dx\\\\ &=-2\int_0^\infty \frac{\sin(|a|tx)}{x}\,dx+2\int_0^\infty \frac{\sin(|a|tx)}{x(x^2+1)}\,dx\\\\ &=-\pi \text{sgn}(t)+2\int_0^\infty \frac{\sin(|a|tx)}{x(x^2+1)}\,dx \end{align}$$

Differentiating once more with respect to $t$, we have for $t\ne 0$, we obtain

$$\begin{align} \frac{\partial^2f(t,a)}{\partial t^2}&=2|a|\int_0^\infty \frac{\cos(|a|tx)}{x^2+1}\,dx\\\\ &=a^2f(t,a) \end{align}$$

Therefore, $f(t,a)$ satisfies the differential equation

$$\frac{\partial^2f(t,a)}{\partial t^2}-a^2f(t,a)=0 \tag 4$$

along with the initial conditions $\displaystyle f(0,a)=\pi/|a|$ and $\displaystyle \left.\frac{\partial f(t,a)}{\partial t}\right|_{t=0^{\pm}}=\mp \pi$.

Solving the differential equation in $(4)$ and enforcing the initial conditions yields

$$\bbox[5px,border:2px solid #C0A000]{f(t,a)=\frac{\pi e^{-|at|}}{|a|}}$$

And we are done!

Answer 2

Herein, we use complex analysis to evaluate the integral of interest. We assume that $t\in \mathbb{R}$ and $a\in \mathbb{R}$. The more general cases for which $t$ and $a$ are complex can be analyzed analogously with special care to account for the signs on the real parts of $t$ and $a$.

Let $f(t,a)$ be represented by the integral

$$\bbox[5px,border:2px solid #C0A000]{f(t,a)=\int_{-\infty}^\infty \frac{e^{ixt}}{x^2+a^2}\,dx }\tag 1$$

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Next we analyze the integral $g(t,a)$ as given by

$$g(t,a)=\oint_C \frac{e^{izt}}{z^2+a^2}\,dz \tag 2$$

where $C$ is the closed contour comprised of $(i)$ the real line segment from $-R$ to $R$ and (ii) the semi-circle centered at the origin with radius $R$ in the upper (lower) half plane for $t>0$ ($t<0$).

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Applying the residue theorem to $(2)$, we find that for $R>|a|$

$$\begin{align} g(t,a)&=2\pi i\,\text{sgn}(t)\, \text{Res}\left(\frac{e^{izt}}{z^2+a^2}, z =i|a|\text{sgn}(t) \right)\\\\ &=\frac{\pi e^{-|at|}}{|a|} \tag3 \end{align}$$

We can write $(2)$

$$g(t,a)=\int_{-R}^R \frac{e^{ixt}}{x^2+a^2}\,dx+\text{sgn}(t)\,\int_0^{\pi\text{sgn}(t)} \frac{e^{itRe^{i\phi}}}{R^2e^{i2\phi}+a^2}\,iRe^{i\phi}\,d\phi \tag 4$$

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As $R\to \infty$, the second integral on the right-hand side of $(4)$ approaches zero while the first integral approaches $f(t,a)$ as given by $(1)$.

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Putting together $(3)$ and $(4)$, we arrive at the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{e^{ixt}}{x^2+a^2}\,dx =\frac{\pi e^{-|at|}}{|a|}} $$

which agrees with the result reported in This Answer, which used only real analysis.