Evaluation of an integral

Calculus (Integrals, Series)
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Tsakanikas Nickos
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Evaluation of an integral

#1

Post by Tsakanikas Nickos »

Evaluate the integral

\[ \displaystyle \int_{-\infty}^{+\infty} \frac{ \cos(\alpha x) }{1+x^2} \mathrm{d}x \]

where \( \alpha \in \mathbb{R} \smallsetminus \{ 0 \} \).
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Tolaso J Kos
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Re: Evaluation of an integral

#2

Post by Tolaso J Kos »

Hello Nickos,

1st way:
Let \( \displaystyle f(a)=\int_{0}^{\infty}\frac{\cos ax}{x^2+1}\,{\rm d}x \). Then :

$$\mathcal{L}\left ( f(a) \right )=\int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos ax}{x^2+1}e^{-as}\,{\rm d}a \,{\rm d}x=\int_{0}^{\infty}\frac{s}{\left ( 1+x^2 \right )\left ( s^2+x^2 \right )}\,{\rm d}x =\frac{\pi}{2(s+1)}$$

Then \( \displaystyle f(a)=\mathcal{L}^{-1}\left \{ \frac{\pi}{2(s+1)} \right \}=\frac{\pi}{2}e^{-\left | a \right |} \). (by distinguishing cases for \( a \). )

Hence your original integral equals twice the integral of \( f(a) \). Hence:

$$\int_{-\infty}^{\infty}\frac{\cos ax}{x^2+1}\,{\rm d}x=\pi e^{-\left | a \right |}$$

2nd way:

Consider the function \( \displaystyle f(z)=\frac{e^{iaz}}{z^2+1} \) which is analytic and has one pole contained in the UHP.
Consider the contour integral \( \displaystyle \oint_{\Gamma}f(z)\,{\rm d}z \) whereas \( \Gamma \) is a semicircle of radius \( R \) in the UHP.

For \( a>0 \) splitting the contour apart we get:
$$\oint_{\Gamma}f(z)\,{\rm d}z =\int_{-R}^{R}f(x)\,{\rm d}x +\oint_{\gamma}f(z)\,{\rm d}z$$

That line integral over the arc vanishes at the upper half plane as \( R \rightarrow + \infty \) since:

$$\begin{aligned}
\left| \oint_{\gamma}f(z) \,{\rm d}z \right| &\overset{z=Re^{i\theta}}{=\! =\! =\! =\!}\left |i R \int_{0}^{\pi} \exp{(i \theta)} \frac{\exp (i a R \exp{(i \theta)})}{ (1+R^2 \exp{(i 2 \theta)})}\,{\rm d}\theta \right|\\
&\leq \frac{1}{R}\int_{0}^{\pi}\exp \left ( -aR \cos \theta \right )\,{\rm d} \theta \\
&\overset{R\rightarrow +\infty}{\longrightarrow} 0
\end{aligned}$$

Hence
$$\int_{-\infty}^{\infty}\frac{\cos ax}{1+x^2}\,{\rm d}x=2\pi i \mathfrak{Res}\left ( f; z=i \right )=\frac{2\pi i e^{-a}}{2i}=\pi e^{-a}, a>0$$

When \( a<0 \) we flip the contour and use the other pole instead. In this case we find that the integral is \( \pi e^a , \;\; a<0\).

Combining the results we get that:

$$\int_{-\infty}^{\infty}\frac{\cos ax}{x^2+1}\,{\rm d}x=\pi e^{-\left | a \right |}, \quad \quad \quad \square$$
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mathofusva
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Re: Evaluation of an integral

#3

Post by mathofusva »

Here is the third way. Without loss of generality, we assume that $\alpha \geq 0$. Let
$$f(\alpha) = \int_0^\infty\,\frac{\cos\alpha x}{1 + x^2}\,dx$$
Applying Leibniz's rule gives
$$f'(\alpha) = -\int_0^\infty\,\frac{x\sin\alpha x}{1 + x^2}.$$
Appealing to
$$\frac{\sin\alpha x}{x(1 + x^2)} = \frac{\sin\alpha x}{x} - \frac{x\sin\alpha x}{1 + x^2},$$
we obtain
$$f'(\alpha) = - \frac{\pi}{2} + \int_0^\infty\,\frac{\sin\alpha x}{x(1 + x^2)}\,dx,$$
and so
$$f''(\alpha) = f(\alpha).$$
It follows that $f(\alpha) = c_1e^{\alpha} + c_2e^{-\alpha}$. The boundedness of $f(\alpha)$ implies $c_1 = 0$, and $f(0) = \pi/2$ yields $c_2 = \frac{\pi}{2}$. So the value of the proposed integral is $2f(\alpha) = \pi\,e^{-\alpha}$.

Remark. For the method of parametric differentiation and integration, see Chapter 19 in the book Excursions in Classical Analysis, by H. Chen, published by MAA, 2010.
Tsakanikas Nickos
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Re: Evaluation of an integral

#4

Post by Tsakanikas Nickos »

Thank you both for your nice solutions!
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