How can I determine whether $\int_1^2\frac{\exp(\sin x)}{x-1}dx$ exists or not?
How to show convergence of an exponential integral
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calculus
integration
improper-integrals
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0Since the region of integration is finite, the only thing that can go wrong is a singularity. The only singularity is at 1 and in a neighborhood of 1 the numerator is bounded below (since $\exp(\sin(1)) > 0$). This integral then converges exactly if $\int_1^{1+\epsilon} \frac{1}{x-1}dx$ converges. – 2012-06-06
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0As a matter of *strategy*, I would suggest you make the substitution $u=x-1$, which changes our integral to $\int_0^1 \frac{\exp(\sin(u+1))}{u}\,du$, which may look more like stuff you have previously done. No mathematical difference, but perhaps still useful. – 2012-06-06