0-pi-tsin-t-1-t-2-dt-

Question Number 210326 by Shrodinger last updated on 06/Aug/24

\int_{0}^{π} \frac{t s i n (t)}{1 + t^{2}} d t

Commented by Spillover last updated on 06/Aug/24

i g o t 1.6583 i s i t r i g h t ?

Commented by Shrodinger last updated on 06/Aug/24

n o s i r 0, 20 \dots

Commented by Ghisom last updated on 07/Aug/24

\approx . 841492

Commented by Frix last updated on 07/Aug/24

\underset{0}{\int^{π}} \frac{t \cos t}{t^{2} + 1} d t + i \underset{0}{\int^{π}} \frac{t \sin t}{t^{2} + 1} d t = \underset{0}{\int^{π}} \frac{{t e}^{i t}}{t^{2} + 1} d t

\Rightarrow

\underset{0}{\int^{π}} \frac{t \sin t}{t^{2} + 1} d t = imag (\underset{0}{\int^{π}} \frac{{t e}^{i t}}{t^{2} + 1} d t)

\dots but I^{'} m not good at this .

Anyway we can get the exact result for

\underset{- \infty}{\int^{+ \infty}} \frac{t \sin t}{t^{2} + 1} d t = \frac{π}{e}

using contour integration .

Commented by mahdipoor last updated on 07/Aug/24

This site can solve many integrals for

example it has solved this integral with the

help of imaginary numbers and ci and si

functions .

www . integral - calculator . com

Commented by Frix last updated on 07/Aug/24

Yes, I also found this site but {there}^{'} s no

useable exact solution, we must approximate

at the end .

Answered by Berbere last updated on 08/Aug/24

\frac{t}{1 + t^{2}} s i n (t) = \frac{1}{2} (\frac{1}{t + i} + \frac{1}{t - i}) s i n (t)

= R e {\int^{π}}_{0} \frac{s i n (t)}{t + i} d t = \int_{i}^{i + π} \frac{s i n (u - i)}{u} d u

= \int_{i}^{i + π} \frac{s i n (u) c o s (i) - c o s (u) s i n (i)}{u} d u; c o s (i) = c h (1)

s i n (i) = i s h (1)

= R e {\int_{i}^{i + u} c h (1) \frac{s i n u}{u} d u + i s h (1) \int_{i}^{i + u} \frac{- c o s (u)}{u} d u}

= R e {S i (i + π) - S i (i)} c h (1) + R e (i s h (1) C i (i) - C i (i + π)}

Commented by hardmath last updated on 11/Aug/24

Dear professor, your solutions are perfect