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I wonder how to solve $$\int_0^\infty\frac{\sin x}{x+1}dx$$I went about solving it like this: \begin{align*}\int_0^\infty\frac{\sin x}{x+1}dx& =\int_1^\infty\frac{\sin(x-1)}xdx=\int_1^\infty\frac{\sin(x)\cos(1)-\cos(x)\sin(1)}xdx\\ & =\cos(1)\left(\frac{\pi}2-\text{Si}(1)\right)-\sin(1)\int_1^\infty\frac{\cos x}xdx\end{align*} So it remains to find how to solve $$\int_1^\infty\frac{\cos x}xdx$$Or at least write it in terms of special functions. Maybe there is a way to solve the original integral without going through this integral.

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    $\begingroup$ Is that not just $-\operatorname{Ci}(1)$? $\endgroup$
    – robjohn
    Sep 15, 2023 at 2:30
  • $\begingroup$ How did you get $\operatorname{Si}(x)$? What is $x$? Your initial integral does not depend on $x$, it is a real number. $\endgroup$
    – Gary
    Sep 15, 2023 at 2:48
  • $\begingroup$ Btw, you can re-write your integral as $$ \int_0^{ + \infty } {\frac{{{\rm e}^{ - x} }}{{1 + x^2 }}{\rm d}x} , $$ which has much better convergence properties for numerical calculations. $\endgroup$
    – Gary
    Sep 15, 2023 at 3:31

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What you have done is absolutely correct. In the last step, you can write $$-\int_{x}^\infty{\frac{\cos t}{t}\mathrm{d}t} = \operatorname{Ci}(x)$$ Where $\operatorname{Ci}(x)$ is the cosine integral of $x$. Otherwise, you can use numerical approximations to get the answer without finding the antiderivative.

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  • $\begingroup$ Thanks. Wolfram Mathworld simply stated that $\text{Ci}(x)=\int\frac{\cos x}x dx$ without giving bounds. $\endgroup$ Sep 15, 2023 at 14:25
  • $\begingroup$ @KamalSaleh Are you referring to an entry other than the one devoted to $\operatorname{Ci}(x)$? The definite integral is listed here (line $(1)$) $\endgroup$
    – user170231
    Sep 15, 2023 at 15:17
  • $\begingroup$ @user170231 Oh I see, that's my fault. I found a picture on the internet and didn't bother clicking on the link. $\endgroup$ Sep 15, 2023 at 15:18

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