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Does the integral $$\int_{0}^{\infty}(\sin\left(1-e^{-x}\right)-\sin\left(1\right))dx$$ converge? If so, how would I solve it? It seems to be a $\infty-\infty$ indeterminate limit but I have been unable to prove it.

I tried finding a antiderivative by using the sum angle identity to write the integral as $$\int_{0}^{\infty}\left[ \sin\left(1\right)\left(\cos\left(e^{-x}\right)-1\right)-\sin\left(e^{-x}\right)\cos\left(1\right)\right]dx$$ which turns it into a linear combination of 2 seemingly convergent integrals, $$\int_{0}^{\infty}\sin\left(e^{-x}\right)dx$$ and $$\left(\cos\left(e^{-x}\right)-1\right).$$ How do I go from here?

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    $\begingroup$ Hint: By the mean value theorem we have $$\vert \sin(1-e^{-x})-\sin(1)\vert \leq (\sup_{y\in \mathbb{R}} \vert \sin'(y)\vert) \vert 1-e^{-x}-1\vert.$$ This gives, after simplifying it, a nice upper bound to work out convergence. $\endgroup$ Commented Oct 26 at 22:05
  • $\begingroup$ Another possible hint: after converting to product it can be estimated by sine integral. $\endgroup$ Commented Oct 26 at 22:09
  • $\begingroup$ $\int_{0}^{\infty}\sin(e^{-x})dx$ and $\int_{0}^{\infty}(\cos(e^{-x})-1)dx$ can be computed with the help of the sine and cosine integral $\endgroup$ Commented Oct 26 at 22:17

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First perform the substitution $u = e^{-x}$. We obtain

\begin{align} I &= \int_{0}^{\infty} (\sin(1-e^{-x}) - 1) \, dx \\ &= \int_{0}^{1} \frac{\sin(1-u) - \sin 1}{u} \, du \\ &= \sin 1 \int_{0}^{1} \frac{\cos u - 1}{u} \, du - \cos 1 \int_{0}^{1} \frac{\sin u}{u} \, du \end{align}

The first integral is well-known to be convergent, it evaluates to $\text{Si}(1)$. The second integral converges after looking at the taylor series of $\cos u$ near $0$. In fact, the integral evaluates to

$$\text{Ci}(1) \sin(1) - \text{Si}(1) \cos(1) - \gamma \sin(1) \approx-0.712965 < \infty$$

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