integration - Evaluate fourier integral : $int_{-infty}^infty frac{e^{isx} ds}{(s-tfrac{7i}{2})^2 + (1/2)^2}$

$$
\mbox{Evaluate}\quad
\int_{-\infty}^{\infty}{\mathrm{e}^{\mathrm{i}sx} \over
\left(\,{s - 7\mathrm{i}/2}\,\right)^{2} + \left(\,{1/2}\,\right)^2}\,\mathrm{d}s
$$

Here I am having trouble as using simply the fourier inverse of $2\left\vert\,{a}\,\right\vert/\left(\,{s^{2} + a^{2}}\,\right)$ which is $\mathrm{e}^{-\left\vert\,{ax}\,\right\vert}$ gives wrong answer as denominator has
$s-7\mathrm{i}/2$.

I can do convolution theorem and then find inverse, but I want to directly evaluate from this expression.

Thank you !.

And if this requires residue theorem, where can I learn this ?.

Answer

Since you are not familiar with complex analysis, we shall resort to using differential equations (these two are, in fact, the most used methods when dealing with integrals that depend on parameters).

Using Lebesgue's dominated convergence theorem, we have that

$$\int _{-\infty} ^\infty \frac {\mathrm e^{\mathrm i s x}} {(s - \frac {7 \mathrm i} 2)^2 + (\frac 1 2)^2} \ \mathrm d s = \lim _{R \to \infty} \int _{-R} ^R \frac {\mathrm e^{\mathrm i s x}} {(s - \frac {7 \mathrm i} 2)^2 + (\frac 1 2)^2} \ \mathrm d s \ .$$

Next, notice that

$$\frac 1 {(s - \frac {7 \mathrm i} 2)^2 + (\frac 1 2)^2} = \frac 1 {(s - \frac {7 \mathrm i} 2)^2 - (\frac {\mathrm i} 2)^2} = \frac 1 {(s - 3 \mathrm i) (s - 4 \mathrm i)} = \frac 1 {\mathrm i} \left( \frac 1 {s - 4 \mathrm i} - \frac 1 {s - 3 \mathrm i} \right) ,$$

therefore your integral becomes

$$\frac 1 {\mathrm i} \lim _{R \to \infty} \int _{-R} ^R \frac {\mathrm e^{\mathrm i s x}} {s - 4 \mathrm i} \ \mathrm d s - \frac 1 {\mathrm i} \lim _{R \to \infty} \int _{-R} ^R \frac {\mathrm e^{\mathrm i s x}} {s - 3 \mathrm i} \ \mathrm d s \ .$$

For fixed $a > 0$ and $R>0$ let

$$I(x) = \int _{-R} ^R \frac {\mathrm e^{\mathrm i s x}} {s - a \mathrm i} \ \mathrm d s \ .$$

Using again Lebesgue's dominated convergence theorem, we may differentiate with respect to $x$ inside the integral, so

$$I'(x) = \int _{-R} ^R \frac {\mathrm i s \ \mathrm e^{\mathrm i s x}} {s - a \mathrm i} \mathrm d s = \int _{-R} ^R \frac {\mathrm i (s - a \mathrm i + a \mathrm i) \ \mathrm e^{\mathrm i s x}} {s - a \mathrm i} \mathrm d s = \mathrm i \int _{-R} ^R \mathrm e ^{\mathrm i s x} \mathrm d s - a \int _{-R} ^R \frac {\mathrm e^{\mathrm i s x}} {s - a \mathrm i} \mathrm d s = 2 \mathrm i \frac {\sin R x} x - a I(x) \ ,$$

which is a nonhomogeneous linear differential equation of order $1$ in $I$, which is easily solvable with the usual method and has for (unique) solution

$$I(x) = 2 \mathrm i \mathrm e ^{-ax} \int _{-\infty} ^x \mathrm e ^{a y} \frac {\sin R y} y \ \mathrm d y = 2 \mathrm i \mathrm e ^{-ax} \int _{-\infty} ^{Rx} \mathrm e ^{\frac {au} R} \frac {\sin u} u \ \mathrm d u \ .$$

For $x>0$ we deduce, using Lebesgue's dominated convergence theorem for the third time (this should convince you about its importance), that

$$\lim _{R \to \infty} \int _{-R} ^R \frac {\mathrm e^{\mathrm i s x}} {s - a \mathrm i} \ \mathrm d s \ = \lim _{R \to \infty} I(x) = 2 \mathrm i \mathrm e ^{-ax} \lim _{R \to \infty} \int _{-\infty} ^{Rx} \mathrm e ^{\frac {au} R} \frac {\sin u} u \ \mathrm d u = 2 \mathrm i \mathrm e ^{-ax} \int _{-\infty} ^\infty \frac {\sin u} u \ \mathrm d u = 2 \pi \mathrm i \ \mathrm e ^{-ax}$$

whence, for $x>0$, the integral that you are looking for is found to be

$$2 \pi (\mathrm e^{-4x} - \mathrm e^{-3x}) \ .$$

(For various methods of computing $\int _{-\infty} ^\infty \frac {\sin u} u \mathrm d u$, you may look here (and at the link therein) and here.)

If $x<0$ the, using the same approach as above, we get that $\lim _{R \to \infty} \int _{-\infty} ^{Rx} = \int _{-\infty} ^{-\infty} = 0$, therefore the desired integral is $0$.

Finally, for $x=0$, we could devise some complicated method to compute the corresponding integral, but it is easier to notice that the inverse Fourier transform will be continuous at $0$, and since the lateral limits are $0$ (using the result obtained so far), then the integral itself must be $0$ for $x=0$.

To conclude, the desired result is $\begin{cases} 2 \pi (\mathrm e^{-4x} - \mathrm e^{-3x}), & x>0 \\ 0, & x \le 0 \end{cases}$.