# Prove that $\int_0^\infty\,\frac{\sin(kx)}{x(x^2+1)}\,\text{d}x=\frac{\pi}{2}\,\left(1-\exp(-k)\right)$ for all $k\in\mathbb{R}_{\ge0}$.

I found the following result (I apologize if this has been posted before, but I could not find anything here). I am wondering whether there is an approach without using contour integration.

Let $k$ be a nonnegative real number. Prove that $$\int_0^\infty\,\frac{\sin(kx)}{x\,\left(x^2+1\right)}\,\text{d}x=\frac{\pi}{2}\,\Big(1-\exp(-k)\Big)=\pi\,\exp\left(-\frac{k}{2}\right)\,\sinh\left(\frac{k}{2}\right)\,.$$

I am especially interested in a solution not using complex analysis. However, a complex-analytic solution that is different from mine is very welcome as well. I have three approaches, but all of them end up using complex analysis as a major part.

We also have this very nice consequence. This result can be proven on its own without knowing the integral $\displaystyle \int_0^\infty\,\frac{\sin(kx)}{x\,\left(x^2+1\right)}\,\text{d}x$.

Let $k$ be a nonnegative real number. Then, $$\int_0^\infty\,\frac{1-\cos(kx)}{x^2\,\left(x^2+1\right)}\,\text{d}x=\frac{\pi}{2}\,\big(k-1+\exp(-k)\big)\,.$$ Equivalently, $$\int_0^\infty\,\frac{\sin^2(kx)}{x^2\,\left(x^2+1\right)}\,\text{d}x=\frac{\pi}{4}\,\big(2k-1+\exp(-2k)\big)\,.$$

Interestingly, during my quest to obtain this integral, I discovered two more integral relations, although I do not know how to obtain the exact values of any of them. The exact values of these integrals involve the exponential integral $\text{Ei}$, where $\text{Ei}(x)=\displaystyle\text{PV}\int_{-\infty}^x\,\frac{\exp(t)}{t}\,\text{d}t$ for all $x\in\mathbb{R}$. Both of the results below also came from contour integrations.

Let $k$ be a nonnegative real number. Then, $$\int_0^\infty\,\frac{1-\cos(kx)}{x\,\left(x^2+1\right)}\,\text{d}x=-\frac{2}{\pi}\,\int_0^\infty\,\frac{\ln(x)\,\sin(kx)}{x\,\left(x^2+1\right)}\,\text{d}x\,.$$ and $$\int_0^\infty\,\frac{\sin(kx)}{x^2+1}\,\text{d}x=-\frac{2}{\pi}\,\int_0^\infty\,\frac{\ln(x)\,\cos(kx)}{x^2+1}\,\text{d}x\,.$$

Mathematica says that $$\int_0^\infty\,\frac{\sin(kx)}{x^2+1}\,\text{d}x=\frac{\exp(-k)\,\text{Ei}(+k)-\exp(+k)\,\text{Ei}(-k)}{2}$$ and that $$\int_0^\infty\,\frac{1-\cos(kx)}{x\,\left(x^2+1\right)}\,\text{d}x=\gamma+\ln(k)-\frac{\exp(-k)\,\text{Ei}(+k)+\exp(+k)\,\text{Ei}(-k)}{2}\,.$$ Here, $\gamma\approx 0.57722$ is the Euler-Mascheroni constant.

Approach I.

Consider the meromorphic function $f(z):=\dfrac{\exp(\text{i}kz)}{z(z^2+1)}$ for all $z\in \mathbb{C}\setminus \{0,-\text{i},+\text{i}\}$. For $\epsilon\in(0,1)$, let $C_\epsilon$ be the positively oriented contour $$\left[+\epsilon,+\frac{1}{\epsilon}\right]\cup \Biggl\{\frac{\exp(\text{i}\theta)}{\epsilon}\,\Bigg|\,\theta\in[0,\pi]\Biggr\}\cup\left[-\frac{1}{\epsilon},-\epsilon\right]\cup \Big\{\epsilon\,\exp(\text{i}\theta)\,\Big|\,\theta\in[\pi,0]\Big\}\,.$$ Write $\Gamma_r$ for the positively-oriented (with respect to $0$) semicircle $\Big\{r\,\exp(\text{i}\theta)\,\Big|\,\theta\in[0,\pi]\Big\}$ for every $r>0$. We have $$I(k):=\lim\limits_{\epsilon\to 0^+}\,\oint_{C_\epsilon}\,f(z)\,\text{d}z=2\pi\text{i}\,\text{Res}_{z=\text{i}}\big(f(z)\big)=-\pi\text{i}\,\exp(-k)\,.$$ Now, note that $$\lim_{\epsilon\to0^+}\,\int_{\Gamma_\epsilon}\,f(z)\,\text{d}z=\pi\text{i}\text{ and }\lim_{\epsilon\to0^+}\,\int_{\Gamma_{\frac{1}{\epsilon}}}\,f(z)\,\text{d}z=0\,.$$ Because $$I(k)=\int_0^\infty\,\frac{\exp(+\text{i}kx)-\exp(-\text{i}kx)}{x(x^2+1)}\,\text{d}x-\lim_{\epsilon\to0^+}\,\int_{\Gamma_\epsilon}\,f(z)\,\text{d}z+\lim_{\epsilon\to0^+}\,\int_{\Gamma_{\frac{1}{\epsilon}}}\,f(z)\,\text{d}z\,,$$ we see that $$I(k)=2\text{i}\,\int_0^\infty\,\frac{\sin(kx)}{x(x^2+1)}\,\text{d}x-\pi\text{i}\,.$$ The result follows immediately.

Approach II.

We apply Richard Feynman's integral trick. First, define $J(k)$ to be the required integral: $$J(k):=\int_0^\infty\,\frac{\sin(kx)}{x(x^2+1)}\,\text{d}x\,.$$ Thus, by the Leibniz Integral Rule, we have $J'(k)=\displaystyle \int_0^\infty\,\frac{\cos(kx)}{x^2+1}\,\text{d}x$. Let $g(z):=\dfrac{\exp(\text{i}kz)}{z^2+1}$ for all $z\in\mathbb{C}\setminus\{-\text{i},+\text{i}\}$. It follows that $$\lim_{\epsilon\to0^+}\,\oint_{C_\epsilon}\,g(z)\,\text{d}z=2\,\int_{0}^\infty\,\frac{\cos(kx)}{x^2+1}\,\text{d}x\,,$$ where $C_\epsilon$ is the positively oriented contour $$\left[+\epsilon,+\frac{1}{\epsilon}\right]\cup \Biggl\{\frac{\exp(\text{i}\theta)}{\epsilon}\,\Bigg|\,\theta\in[0,\pi]\Biggr\}\cup\left[-\frac{1}{\epsilon},-\epsilon\right]\cup \Big\{\epsilon\,\exp(\text{i}\theta)\,\Big|\,\theta\in[\pi,0]\Big\}\text{ for }\epsilon\in(0,1)\,.$$ Furthermore, we have $$\oint_{C_\epsilon}\,g(z)\,\text{d}z=2\pi\text{i}\,\text{Res}_{z=\text{i}}\big(g(z)\big)=\pi\,\exp(-k)\text{ for all }\epsilon\in(0,1)$$ Ergo, $J'(k)=\displaystyle\int_{0}^\infty\,\frac{\cos(kx)}{x^2+1}\,\text{d}x=\dfrac{\pi}{2}\,\exp(-k)$. Since $J(0)=0$, $$J(k)=\int_0^k\,J'(t)\,\text{d}t=\frac{\pi}{2}\,\int_0^k\,\exp(-t)\,\text{d}t=\frac{\pi}{2}\,\big(1-\exp(-k)\big)\,.$$

Approach III.

It is easy to see that $\dfrac{\sin(t)}{t}=\displaystyle\frac{1}{2}\,\int_{-1}^{+1}\,\exp(\text{i}t\tau)\,\text{d}\tau$ for all $t\neq 0$. That is, the required integral is given by \begin{align}J(k):=\int_0^\infty\,\frac{\sin(kx)}{x\,\left(x^2+1\right)}\,\text{d}x&=\frac{1}{2}\,\int_{0}^\infty\,\frac{k}{x^2+1}\,\int_{-1}^{+1}\,\exp(\text{i}kxt)\,\text{d}t\,\text{d}x\\&=\frac{1}{2}\,\int_{-\infty}^{+\infty}\,\frac{k}{x^2+1}\,\int_{0}^{1}\,\exp(\text{i}kxt)\,\text{d}t\,\text{d}x\,.\end{align} Using Fubini's Theorem, we obtain $$J(k)=\frac{1}{2}\,\int_0^1\,k\,\int_{-\infty}^{+\infty}\,\frac{\exp(\text{i}kxt)}{x^2+1}\,\text{d}x\,\text{d}t\,.$$
For a real number $R>1$, let $\gamma_R$ be the positively oriented contour $$[-R,+R]\cup\big\{R\,\exp(\text{i}\theta)\,\big|\,\theta\in[0,2\pi]\big\}\,.$$ Then, for $\omega \geq 0$, we have $$\lim_{R\to\infty}\,\oint_{\gamma_R}\,\frac{\exp(\text{i}\omega z)}{z^2+1}\,\text{d}z=\int_{-\infty}^{+\infty}\,\frac{\exp(\text{i}\omega x)}{x^2+1}\,\text{d}x=:K(\omega)\,.$$ Ergo, $$K(\omega)=2\pi\text{i}\,\text{Res}_{z=\text{i}}\left(\frac{\exp(\text{i}\omega z)}{z^2+1}\right)=2\pi\text{i}\,\left(\frac{\exp(-\omega)}{2\text{i}}\right)=\pi\,\exp(-\omega)\,.$$ As $\displaystyle J(k)=\frac{1}{2}\,\int_0^1\,k\,K(kt)\,\text{d}t$, we conclude that $$J(k)=\frac{\pi}{2}\,\int_0^1\,k\,\exp(-kt)\,\text{d}t=\frac{\pi}{2}\,\big(1-\exp(-k)\big)\,.$$ From this proof, we also obtain $$\int_0^\infty\,\frac{\cos(kx)}{x^2+1}\,\text{d}x=\frac{1}{2}\,K(k)=\frac{\pi}{2}\,\exp(-k)\,.$$

P.S. I put my solutions in spoilers for people who want to try to solve the problem without being led on by my attempts. If you do not want to waste your time doing what I have already done, then please look at the spoilers. All of the solutions in the spoilers are complex-analytic proofs anyhow, so if you are giving me a real-analytic proof, then there is no way you are going to replicate my work.

• "I am wondering whether there is an approach without using contour integration." > tags contour-integration – Kenny Lau Aug 2 '18 at 19:20
• @KennyLau Because my solutions involve contour integration. – Batominovski Aug 2 '18 at 19:20
• Spoilers aren't helpful at all in a question. After all, you are looking for solutions other than the approaches you've taken, right? So why play games and hide them behind spoilers? – amWhy Aug 2 '18 at 19:21
• @amWhy Because some people might not want to see a solution. In cases like this, seeing a solution can corrupt one's imagination. – Batominovski Aug 2 '18 at 19:22
• @amWhy The spoilers are for someone who wants to work on this problem without knowing a solution beforehand. I like attacking problems on my own, and believe many people like that too. If you don't want to waste your time, you are welcome to look at the spoilers. – Batominovski Aug 2 '18 at 19:25

HINT:

Let $f(k)$ be given by the integral

$$f(k) =\int_0^\infty \frac{\sin(kx)}{x(x^2+1)}\,dx \tag1$$

Inasmuch as the improper integral $\int_0^\infty \frac{x\sin(kx)}{x^2+1}\,dx$ converges uniformly for $|k|\ge \delta>0$, we can differentiate twice under the integral in $(1)$ to reveal

$$f''(k)-f(k)=-\frac\pi2 \text{sgn}(k)\tag2$$

Solve $(2)$ subject to the initial conditions $f(0)=0$ and $f'(0)=\frac\pi2$.

NOTE:

$$\int_0^\infty \frac{x\sin(kx)}{x^2+1}\,dx=\int_0^\infty \frac{(x^2+1-1)\sin(kx)}{x(x^2+1)}\,dx=\int_0^\infty \frac{\sin(kx)}{x}\,dx-\int_0^\infty \frac{\sin(kx)}{x(x^2+1)}\,dx$$

• I will accept your answer eventually. I just want to wait a bit more. I want to see how many more solutions I can get. By the end of this week, I will accept the answer. – Batominovski Aug 7 '18 at 16:10
• @Batominovski Much appreciative. – Mark Viola Aug 7 '18 at 16:15
• Many thanks for this great answer by the way. :D – Batominovski Aug 7 '18 at 16:16
• You're welcome. It was a pleasure. – Mark Viola Aug 7 '18 at 16:17

You can use a nice property of the Laplace transform to computer your integral: $$\int_0^{\infty} f(x) g(x)\,dx=\int_0^{\infty} \mathcal{L}\{f(x)\}(s)\,\mathcal{L}^{-1}\{g(x)\}(s) \,ds$$ In your case, letting $f(x)=\sin (kx)$ and $g(x)=\frac{1}{x(x^2+1)}$, we can show that \begin{align} \int_0^{\infty} \frac{\sin(kx)}{x(x^2+1)}\,dx &= k\int_0^{\infty} \frac{1-\cos(s)}{k^2+s^2}\,ds \\ &=\frac{\pi}{2}-k\int_0^{\infty} \frac{\cos(s)}{k^2+s^2}\,ds \\ &=\frac{\pi}{2}-\frac{k}{2}\int_{-\infty}^{\infty} \frac{e^{is}}{k^2+s^2}\,ds \\ &=\frac{\pi}{2} - \frac{k}{2} \sqrt{2\pi} \,\mathcal{F}\left\{\frac{1}{k^2+s^2}\right\}(\omega)\Biggr|_{\omega=1} \\ &= \frac{\pi}{2} - \frac{k}{2}\sqrt{2\pi} \sqrt{\frac{\pi}{2}} \frac{e^{-k}}{k} \\ &=\frac{\pi}{2}-\frac{\pi}{2}e^{-k} \\ &= \frac{\pi}{2}\left(1-e^{-k}\right) \end{align} Where we used the Fourier Transform. Thus, $$\int_0^{\infty} \frac{\sin(kx)}{x(x^2+1)}\,dx=\frac{\pi}{2}\left(1-e^{-k}\right) \quad \text{for } k\in\mathbb{R}^+$$

• If one is going to appeal to a FT table to arrive at a result, then why bother to begin with applying the LT integration by parts? Simply write $$\int_0^\infty \frac{\sin(kx)}{x(x^2+1)}\,dx=\int_0^k \int_0^\infty \frac{\cos(k'x)}{x^2+1}\,dx\,dk'$$ – Mark Viola Aug 6 '18 at 14:37