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

$${\int }_0^{\pi }\frac{tsin\left(t\right)}{1+t^2}dt$$

Commented by Spillover last updated on 06/Aug/24

$$igot1.6583isitright?$$

Commented by Shrodinger last updated on 06/Aug/24

$$nosir0,20...$$

Commented by Ghisom last updated on 07/Aug/24

$$\approx .841492$$

Commented by Frix last updated on 07/Aug/24

$$\underset{0}{\stackrel{\pi }{\int }}\frac{t\cos t}{t^2+1}dt+\mathrm{i}\underset{0}{\stackrel{\pi }{\int }}\frac{t\sin t}{t^2+1}dt=\underset{0}{\stackrel{\pi }{\int }}\frac{{te}^{it}}{t^2+1}dt$$$$\Rightarrow$$$$\underset{0}{\stackrel{\pi }{\int }}\frac{t\sin t}{t^2+1}dt=\mathrm{imag}\left(\underset{0}{\stackrel{\pi }{\int }}\frac{{te}^{it}}{t^2+1}dt\right)$$$$...\mathrm{but}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{good}\mathrm{at}\mathrm{this}.$$$$$$$$\mathrm{Anyway}\mathrm{we}\mathrm{can}\mathrm{get}\mathrm{the}\mathrm{exact}\mathrm{result}\mathrm{for}$$$$\underset{-\mathrm{\infty }}{\stackrel{+\mathrm{\infty }}{\int }}\frac{t\sin t}{t^2+1}dt=\frac{\pi }{\mathrm{e}}$$$$\mathrm{using}\mathrm{contour}\mathrm{integration}.$$

Commented by mahdipoor last updated on 07/Aug/24

$$$$$$\mathrm{This}\mathrm{site}\mathrm{can}\mathrm{solve}\mathrm{many}\mathrm{integrals}\mathrm{for}$$$$\mathrm{example}\mathrm{it}\mathrm{has}\mathrm{solved}\mathrm{this}\mathrm{integral}\mathrm{with}\mathrm{the}$$$$\mathrm{help}\mathrm{of}\mathrm{imaginary}\mathrm{numbers}\mathrm{and}\mathrm{ci}\mathrm{and}\mathrm{si}$$$$\mathrm{functions}.$$$$\mathrm{www}.\mathrm{integral}-\mathrm{calculator}.\mathrm{com}$$

Commented by Frix last updated on 07/Aug/24

$$\mathrm{Yes},\mathrm{I}\mathrm{also}\mathrm{found}\mathrm{this}\mathrm{site}\mathrm{but}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}$$$$\mathrm{useable}\mathrm{exact}\mathrm{solution},\mathrm{we}\mathrm{must}\mathrm{approximate}$$$$\mathrm{at}\mathrm{the}\mathrm{end}.$$

Answered by Berbere last updated on 08/Aug/24

$$\frac{t}{1+t^2}sin\left(t\right)=\frac{1}{2}(\frac{1}{t+i}+\frac{1}{t-i})sin\left(t\right)$$$$=Re{\stackrel{\pi }{\int }}_0\frac{sin\left(t\right)}{t+i}dt={\int }_i^{i+\pi }\frac{sin(u-i)}{u}du$$$$={\int }_i^{i+\pi }\frac{sin\left(u\right)cos\left(i\right)-cos\left(u\right)sin\left(i\right)}{u}du;cos\left(i\right)=ch\left(1\right)$$$$sin\left(i\right)=ish\left(1\right)$$$$=Re\{{\int }_i^{i+u}ch\left(1\right)\frac{sinu}{u}du+ish\left(1\right){\int }_i^{i+u}\frac{-cos\left(u\right)}{u}du\}$$$$=Re\{Si(i+\pi )-Si\left(i\right)\}ch\left(1\right)+Re(ish\left(1\right)Ci\left(i\right)-Ci(i+\pi )\}$$

Commented by hardmath last updated on 11/Aug/24

$$$$Dear professor, your solutions are perfect

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