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I was wandering if it possible to evaluate the value of the following improper integral: $$ \int_0^1 \sin\left(\frac{1}{t}\right)\,dt $$

It is convergent since $\displaystyle\int_0^1 \left|\sin\left(\frac{1}{t}\right)\right|\,dt\leq \int_0^1 \;dt$, but I don't know if it is possible to calculate its value.

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Substituting $t = 1/x$ shows that the integral equals $$ I=\int_1^\infty \frac{\sin x}{x^2} dx. $$ Using integration by parts or otherwise, the indefinite integral can be shown to be $$ \text{Ci}(x)-\frac{\sin (x)}{x}, $$ where $$ \text{Ci}(x) = -\int_x^\infty \frac{\cos y}{y} dy $$ is the Cosine Integral.

Plugging in the limits gives $$ I = \sin 1 - \text{Ci}(1). $$

Unfortunately this is not elementary.

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  • $\begingroup$ The difference of two non elementary values is not necessarily non elementary (although it probably is) unless by elementary you are referring to a level of education. $\endgroup$ – DanielV Dec 13 '14 at 19:30
  • $\begingroup$ @DanielV I would consider $\sin 1$ to be elementary and the other term nonelementary. $\endgroup$ – user111187 Dec 13 '14 at 21:53
  • $\begingroup$ +1. Pretty short. We don't have "to reinvent the wheel": $\,{\rm Ci}$ is fine enough !!!. $\endgroup$ – Felix Marin Dec 14 '14 at 20:43
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Since, through a change of variable and integration by parts: $$\int_{0}^{1}\sin\left(\frac{1}{t}\right)\,dt = \int_{1}^{+\infty}\frac{\sin t-\sin 1}{t^2}\,dt + \sin 1 = \sin 1+\int_{1}^{+\infty}\frac{\cos t}{t}\,dt$$ we just need to compute: $$\int_{1}^{+\infty}\frac{\cos t}{t}\,dt = \sum_{k=0}^{+\infty}\int_{1+k\pi}^{1+(k+1)\pi}\frac{\cos t}{t}\,dt=\int_{1}^{\pi+1}\cos t\sum_{k=0}^{+\infty}\frac{(-1)^k}{t+k\pi}\,dt $$ that is: $$\frac{1}{2\pi}\int_{1}^{\pi+1}\cos(t)\left(\psi\left(\frac{\pi+t}{2\pi}\right)-\psi\left(\frac{t}{2\pi}\right)\right)\,dt $$ and we are left with the integral of a smooth function over a compact interval.

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