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How can I determine whether $\int_1^2\frac{\exp(\sin x)}{x-1}dx$ exists or not?

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    As 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

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The point of trouble is at the $x=1$ end where the denominator goes to $0$. Near this point, the numerator is just some nonzero continuous function, so for small $\varepsilon$ you can approximate $ \int_1^{1+\varepsilon} \frac{e^{\sin x}}{x-1} dx \approx \int_1^{1+\varepsilon} \frac{e^{\sin 1}}{x-1} dx \approx e^{\sin 1} \int_0^{\varepsilon} u^{-1} du $ which doesn't look like it exists.