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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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Since 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. –  Chris Janjigian Jun 6 '12 at 14:29
    
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. –  André Nicolas Jun 6 '12 at 15:12

1 Answer 1

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.

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