# Prove that $\forall k>0,\ \int_0^{+\infty}\frac{\sin x}{x+k}dx<\frac1k$.

Let $k>0$.

Try1: \begin{align} \int_0^{+\infty}\frac{\sin x}{x+k}dx&=\int_k^{+\infty}\frac{\sin (t-k)}{t}dt\tag{$t=x+k$}\\ &=\int_k^{+\infty}\frac{\sin t\cos k-\sin k\cos t}{t}dt\\ &=\cos k\int_k^{+\infty}\frac{\sin t}{t}dt-\sin k\int_k^{+\infty}\frac{\cos t}{t}dt\\ &=\cdots \end{align} Maybe then we can study this integral as a function of $k$ with the help of $\operatorname{Si}$ and $\operatorname{Ci}$? I gave up.

Try2:

We define $$f:z\mapsto\frac{e^{iz}}{z+k}.$$ Let $R>0$, we have $$0=\int_0^{R}\frac{e^{ix}}{x+k}dx+\int_0^{\frac{\pi}2}\frac{e^{iRe^{i\theta}}}{Re^{i\theta}+k}iRe^{i\theta}d\theta-\int_0^{R}\frac{ie^{-x}}{ix+k}dx.$$ For all $\theta\in(0,\frac{\pi}2)$, \begin{align} \left|\frac{e^{iRe^{i\theta}}}{Re^{i\theta}+k}iRe^{i\theta}\right|&=\left|\frac{Re^{iR\cos\theta-R\sin\theta}}{Re^{i\theta}+k}\right|\\ &=\left|\frac{Re^{-R\sin\theta}}{Re^{i\theta}+k}\right|\\ &\underset{R\to+\infty}{\to}0, \end{align} thus, $$\int_0^{+\infty}\frac{e^{ix}}{x+k}dx=\int_0^{+\infty}\frac{ie^{-x}}{ix+k}dx=\int_0^{+\infty}\frac{e^{-x}(x+ik)}{x^2+k^2}dx.$$ Finally, \begin{align} \int_0^{+\infty}\frac{\sin x}{x+k}dx&=\int_0^{+\infty}\frac{ke^{-x}}{x^2+k^2}dx\\ &<\frac1k\int_0^{+\infty}e^{-x}dx\\ &=\frac1k \end{align}

So here are my questions:

1. Is there a simpler/smarter proof (without using the residue theorem)?
2. Can we express the integral as a series of $k$ (distinguishing cases of $k<1$ and $k>1$)?

A wiki page which may help.

• I deleted my answer because I noticed the request for no contour integration. However, I did get the asymptotic bounds $$\sum_{j=0}^{2n+1}(-1)^j\frac{(2j)!}{k^{2j+1}} \le\int_0^\infty\frac{\sin(x)}{k+x}\,\mathrm{d}x \le\sum_{j=0}^{2n}(-1)^j\frac{(2j)!}{k^{2j+1}}$$ which, for $n=0$ gives $$\frac1k-\frac2{k^3} \le\int_0^\infty\frac{\sin(x)}{k+x}\,\mathrm{d}x \le\frac1k$$ – robjohn Feb 25 '18 at 12:15

Here is a simple trick. By integration by parts,

\begin{align*} \int_{0}^{\infty} \frac{\sin x}{x+k} \, dx &= \underbrace{\left[ \frac{1-\cos x}{x+k} \right]_{0}^{\infty}}_{=0} + \int_{0}^{\infty} \frac{1-\cos x}{(x+k)^2} \, dx \\ &= \underbrace{\left[ \frac{x-\sin x}{(x+k)^2} \right]_{0}^{\infty}}_{=0} + \int_{0}^{\infty} \frac{2(x-\sin x)}{(x+k)^3} \, dx \end{align*}

Now if we write $a_n = \int_{0}^{\pi} \frac{\sin x}{(x+k+n\pi)^3} \, dx$ for $n \geq 0$, then $a_n$ is positive, strictly decreasing and

$$\int_{0}^{\infty} \frac{\sin x}{(x+k)^3} \, dx = \sum_{n=0}^{\infty} (-1)^n a_n$$

So by the alternating series test, we know that $\int_{0}^{\infty} \frac{\sin x}{(x+k)^3} \, dx \geq a_0 - a_1 > 0$. Therefore

$$\int_{0}^{\infty} \frac{\sin x}{x+k} \, dx < \int_{0}^{\infty} \frac{2x}{(x+k)^3} \, dx = \frac{1}{k}.$$

Using the formula $$\int_{0}^{\infty}f\left(x\right)g\left(x\right)dx=\int_{0}^{\infty}\left(\mathcal{L}f\right)\left(s\right)\left(\mathcal{L}^{-1}g\right)\left(s\right)ds$$ where $\mathcal{L}\left(\cdot\right)$ and $\mathcal{L}^{-1}\left(\cdot\right)$ are the Laplace and the inverse Laplace transform, respectively, we get $$\int_{0}^{\infty}\frac{\sin\left(x\right)}{x+k}dx=\int_{0}^{\infty}\frac{e^{-kx}}{1+x^{2}}dx<\int_{0}^{\infty}e^{-kx}dx=\color{red}{\frac{1}{k}}.$$

A followup to Marco's answer (+1 to him). The idea of exploiting $\mathcal{L},\mathcal{L}^{-1}$ is slick since it removes the oscillations from the integrand function and makes any numerical approximation way easier. Once we have $$I(k)=\int_{0}^{+\infty}\frac{\sin(x)}{x+k}\,dx = \int_{0}^{+\infty}\frac{e^{-kx}}{1+x^2}\,dx=\int_{0}^{\pi/2}e^{-k\tan\theta}\,d\theta$$ we may also state $$k\,I(k)=\int_{0}^{+\infty}\frac{k e^{-kx}}{1+x^2}\,dx\stackrel{\text{IBP}}{=} 1-\int_{0}^{+\infty}\frac{2x e^{-kx}}{(1+x^2)^2}\,dx$$ where $\int_{0}^{+\infty}\frac{2x e^{-kx}}{(1+x^2)^2}\,dx$ is trivially bounded between $0$ and $\int_{0}^{+\infty}2x e^{-kx}\,dx = \frac{2}{k^2}$.
The same technique also allows to prove $I(k)>\frac{1}{k+1}$ for any $k>0$.