How do you show that
∫∞0dttecos(t)sin(sin(t))=π2(e−1)
I managed to get the left-hand side to equal the imaginary part ofI=∞∫0dtteeitBut I’m not very sure what to do next. I’m thinking of a substitute t↦eit, but I’m not very sure how to evaluate the limit as t→∞. I also tried contour integration, but I’m not exactly sure what contour to draw.
Answer
ecostsinsint=Imexp(eit)=Im∑n≥0enitn!=∑n≥1sin(nt)n!
and since for any a>0 we have ∫+∞0sin(at)tdt=π2 it follows that
∫+∞0ecostsinsintdtt=π2∑n≥11n!=π2(e−1),
pretty simple.
I have a counter-proposal:
∫+∞0(ecostsinsint)2dtt2=π2∑m,n≥1min
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