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

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 $k>0$.