Let $$I_t=\int_0^t|e^{-x}\sin(x)|dx.$$ I am interested in finding $$I_{+\infty}=\lim_{t\to+\infty}I_t.$$ A possibility related integral is $$I_{+\infty}^\prime=\lim_{k\to+\infty}I_k^\prime=\lim_{k\to+\infty}\int_0^ke^{-x}\sin(x)dx=\lim_{k\to+\infty}\left(\frac{1}{2}\left(1-e^{-k}\left(\sin(k)+\cos(k)\right)\right)\right)=\frac{1}{2}.$$ I do not know how to evaluate $I_{+\infty}.$ However, I tried Wolfram Alpha, and according to it, $I_{+\infty}=\frac{1+e^\pi}{2e^\pi-2}.$ How to arrive at this result?
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$\begingroup$ you should be able to evaluate $\int_0^k e^{-x}\sin(x)\mathrm{d}x$ by partially integrating twice and solving the resulting equation $\endgroup$– student91Feb 17 at 16:05
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$\begingroup$ @student91, yes, and I did that in my post. My question is about $I_{+\infty},$ and not $I_{+\infty}^\prime.$ Thanks, though. $\endgroup$– aqualubixFeb 17 at 16:07
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$\begingroup$ Oops, I was confused by the "A possibly related integral is ..., I do not know how to evaluate this integral", have a nice day! $\endgroup$– student91Feb 17 at 16:14
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1$\begingroup$ @AnneBauval, it does. When I looked up this question on MSE, I didn't put the absolute value sign on $\sin(x)$ alone, and hence the linked question didn't show up. Thanks. $\endgroup$– aqualubixFeb 17 at 16:16
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$\begingroup$ @student91, you're right, that is confusing. I'll edit that out so future viewers don't get confused. $\endgroup$– aqualubixFeb 17 at 16:17
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