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I am currently working on an definite integral that requires the following definite integral to be evaluated.

$$ I = \int_{0}^{\infty} \frac{\sin(kx)}{x\left(x^2 + 1\right)} \:dx$$

Where $k \in \mathbb{R}^{+}$

I was wondering what methods can be employed to solve this integral?

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  • $\begingroup$ The symmetry of the integrand suggests that contour integration might be a good option. $\endgroup$ – Travis Nov 2 '18 at 2:28
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The method I took was:

Let

$$ I(t) = \int_{0}^{\infty} \frac{\sin(kxt)}{x\left(x^2 + 1\right)} \:dx$$

Take the Laplace Transform

\begin{align} \mathscr{L} \left[I(t) \right]&= \int_{0}^{\infty} \frac{\mathscr{L} \left[\sin(kxt)\right]}{x\left(x^2 + 1\right)} \:dx \\ &= \int_{0}^{\infty} \frac{kx}{\left(k^2x^2 + s^2\right)x\left(x^2 + 1\right)} \:dx \\ &= \int_{0}^{\infty} \frac{k}{\left(k^2x^2 + s^2\right)\left(x^2 + 1\right)} \:dx \\ &= \int_{0}^{\infty} \left[\frac{k^3}{\left(k^2 - s^2\right)\left(k^2x^2 + s^2 \right)} - \frac{k}{\left(k^2 - s^2\right)\left(x^2 + 1\right)}\right] \:dx \\ &= \left[\frac{k^3}{\left(k^2 - s^2\right)}\frac{\arctan\left(kx\right)}{ks} - \frac{k}{\left(k^2 - s^2\right)}\arctan(x)\right]_{0}^{\infty} \\ &= \frac{k^2}{s\left(k^2 - s^2\right)}\frac{\pi}{2} - \frac{k}{\left(k^2 - s^2\right)}\frac{\pi}{2} \\ &= \frac{k}{k^2 - s^2}\left[ \frac{k}{s} - 1 \right]\frac{\pi}{2} \\ &= \frac{k}{s\left(k + s\right)}\frac{\pi}{2} \end{align}

And thus,

$$ I(t) = \mathscr{L}^{-1}\left[\frac{k}{s\left(k + s\right)}\frac{\pi}{2} \right] = \left[1 - e^{-kt} \right]\frac{\pi}{2}$$

Lastly,

$$ I = I(1) = \left[1 - e^{-k} \right]\frac{\pi}{2}$$

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    $\begingroup$ Looks like it works to me (+1) $\endgroup$ – clathratus Nov 2 '18 at 3:00
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Let's take the advice from the comment section and evaluate the integral using contour integration! Consider the function$$f(z)=\frac {e^{iaz}}{z(1+z^2)}$$And integrate it about the semi-circular contour situated in the upper-half of the complex plane with a small semi circular detour around the origin. Let the larger circle have a radius of $R$ while the smaller detour $\epsilon$ (Picture coming soon). Therefore, calling the contour $\mathrm C$, we have that$$\oint\limits_{\mathrm C}\mathrm dz\, f(z)=\int\limits_{-R}^{-\epsilon}\mathrm dx\, f(x)+\int\limits_{\gamma_{\epsilon}}\mathrm dz\, f(z)+\int\limits_{\epsilon}^{R}\mathrm dx\, f(x)+\int\limits_{\Gamma_{R}}\mathrm dz\, f(z)$$The arc integrals obtain different values depending on the radius. The outer arc integral vanishes per the Estimation Lemma as$$\left|\,\int\limits_{\Gamma_{R}}\mathrm dz\, f(z)\,\right|\leq\frac {\pi R}{R(R^2-1)}\xrightarrow{R\,\to\,\infty}0$$Meanwhile$$\left|\,\int\limits_{\epsilon_{\epsilon}}\mathrm dz\, f(z)\,\right|=-i\int\limits_0^{\pi}\mathrm d\varphi\,\frac {e^{ia\epsilon e^{i\varphi}}}{1+\epsilon^2e^{2i\varphi}}\xrightarrow{\epsilon\,\to\,0}-\pi i$$The contour integral is equal to $2\pi i$ times the sum of the residues. There is only one pole inside our contour at $z=i$. The corresponding residue is calculated as$$\operatorname*{Res}_{z\, =\, i}\frac {e^{iaz}}{z(1+z^2)}=\lim\limits_{z\,\to\, i}\frac {(z-i)e^{iaz}}{z(1+z^2)}=-\frac 1{2e^a}$$Hence$$\oint\limits_{\mathrm C}\mathrm dz\, f(z)\color{red}{=-\pi i e^{-a}}$$Substituting all the values back into the original equation gives$$\int\limits_{-\infty}^{\infty}\mathrm dx\, f(x)=\pi i(1-e^{-a})$$Take the imaginary part of the integral to get$$\int\limits_0^{\infty}\mathrm dx\,\frac {\sin ax}{x(1+x^2)}\color{blue}{=\frac \pi2\left(1-\frac 1{e^a}\right)}$$

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  • $\begingroup$ Thanks for your response. I am deficient in my knowledge of contour integration, so will be reading into it a lot now. Again, thanks. $\endgroup$ – user150203 Nov 2 '18 at 4:54
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You can also use differentiation under the integral sign. Let $$ f \colon \mathbb{R} \to \mathbb{R} \, , \, f(k) = \int \limits_0^\infty \frac{\sin(k x)}{x (1+x^2)} \, \mathrm{d} x \, . $$ Clearly, $f(0) = 0$ . The dominated convergence theorem yields $f \in C^1 (\mathbb{R})$ and $$ f'(k) = \int \limits_0^\infty \frac{\cos(k x)}{1+x^2} \, \mathrm{d} x \, , $$ so $f'(0) = \frac{\pi}{2}$ . This integral has already been discussed on this site many times, but in order to keep the answer self-contained we can just differentiate one more time. As suggested here, for $k > 0$ we let $k x = t$ to find $$ f'(k) = \int \limits_0^\infty \frac{k \cos(t)}{k^2 + t^2} \, \mathrm{d} t \, . $$ Thus we obtain $f \in C^2 (\mathbb{R}^+)$ and \begin{align} f''(k) &= - \int \limits_0^\infty \frac{k^2 - t^2}{(k^2+t^2)^2} \cos(t) \, \mathrm{d} t = - \int \limits_0^\infty \frac{t \sin(t)}{k^2+t^2} \, \, \mathrm{d} t = - \int \limits_0^\infty \frac{x \sin(k x)}{1+x^2} \, \mathrm{d} x \\ &= \int \limits_0^\infty \frac{\sin(k x)}{x (1+x^2)} \, \mathrm{d} x - \int \limits_0^\infty \frac{\sin(k x)}{x} \, \mathrm{d} x = f(k) - \frac{\pi}{2} \, , \, k > 0 \, , \end{align} using the dominated convergence theorem, integration by parts and the known value of the Dirichlet integral. We conclude that $f \rvert_{[0,\infty)}$ is the (unique) solution of the initial value problem $$ \begin{cases} -g'' + g = \frac{\pi}{2} ~~ \mathrm{in} ~ (0,\infty) \\ g(0) = 0 \\ g'(0) = \frac{\pi}{2}\end{cases} \, ,$$ so $f(k) = \frac{\pi}{2}(1-\mathrm{e}^{-k}) \, , \, k \geq 0$ . Since $f$ is obviously antisymmetric, the correct continuation to negative values is given by $$ f(k) = \frac{\pi}{2} \operatorname{sgn}(k) (1-\mathrm{e}^{-|k|}) \, , \, k \in \mathbb{R} \, . $$

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