Calculating the integral $\int_{0}^{\infty} \frac{\cos x}{1+x^2}\mathrm{d}x$ without using complex analysis

Question

Suppose that we do not know anything about the complex analysis (numbers). In this case, how to calculate the following integral in closed form? $$\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x$$

Answer 1

This can be done by the useful technique of differentiating under the integral sign.

In fact, this is exercise 10.23 in the second edition of "Mathematical Analysis" by Tom Apostol.

Here is the brief sketch (as laid out in the exercise itself).

Let $$ F(y) = \int\limits_{0}^{\infty} \frac{\sin xy}{x(1+x^2)} \ dx \ \ \text{for} \quad\quad y > 0$$

Show that

$\displaystyle F''(y) - F(y) + \pi/2 = 0$ and hence deduce that $\displaystyle F(y) = \frac{\pi(1-e^{-y})}{2}$.

Use this to deduce that for $y > 0$ and $a > 0$

$$\displaystyle \int_{0}^{\infty} \frac{\sin xy}{x(x^2 + a^2)} \ dx = \frac{\pi(1-e^{-ay})}{2a^2}$$

and

$$\int_{0}^{\infty} \frac{\cos xy}{x^2 + a^2} dx = \frac{\pi e^{-ay}}{2a}$$

Answer 2

Since $$\frac{x}{1+x^2}=\int_{0}^{\infty}e^{-y}\sin (xy)dy,$$ we have that $$I=\int_{0}^{\infty}\frac{\cos bx}{1+x^2}dx=\int_{0}^{\infty}\frac{\cos bx}{x}dx\int_{0}^{\infty}e^{-y}\sin (xy)dy.$$ Changing the order of integration (which can be justified by the standard method) yields $$I=\int_{0}^{\infty}e^{-y}dy\int_{0}^{\infty}\frac{\sin xy}{x} \cos bx dx.$$ The calculation of the integral (a.k.a. the discontinuous Dirichlet factor) $$\int_{0}^{\infty}\frac{\sin xy}{x} \cos bx dx=\begin{cases}0, & 0 < y < b \\ \ \\ \pi/2, & 0 < b < y, \end{cases}$$ can be easily reduced to the calculation of the standard Dirichlet integral. Therefore, $$I=\frac{\pi}{2}\int_{b}^{\infty}e^{-y}dy=\frac{\pi}{2}e^{-b}.$$

Answer 3

The previous answer is not correct. If you use the Taylor expansion of cosine and integrate termwise you consider integrals of the following form: \begin{eqnarray} \int_{0}^{\infty} \frac{x^{a} \ dx}{1 + x^{2}} = \tfrac{\pi}{2} \sec (\tfrac{\pi a}{2}) \end{eqnarray} which is only well-defined if $-1 < a < 1$.

Answer 4

You can try using the Taylor expansion of $\cos$.