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:
- Is there a simpler/smarter proof (without using the residue theorem)?
- Can we express the integral as a series of $k$ (distinguishing cases of $k<1$ and $k>1$)?
A wiki page which may help.