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

∫_0 ^π ((tsin(t))/(1+t^2 ))dt

$$\int_{\mathrm{0}} ^{\pi} \:\frac{{tsin}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$

Commented by Spillover last updated on 06/Aug/24

i got 1.6583 is it right?

$${i}\:{got}\:\mathrm{1}.\mathrm{6583}\:\:{is}\:{it}\:{right}? \\ $$

Commented by Shrodinger last updated on 06/Aug/24

no sir 0,20...

$${no}\:{sir}\:\mathrm{0},\mathrm{20}... \\ $$

Commented by Ghisom last updated on 07/Aug/24

≈.841492

$$\approx.\mathrm{841492} \\ $$

Commented by Frix last updated on 07/Aug/24

∫_0 ^π ((tcos t)/(t^2 +1))dt+i∫_0 ^π ((tsin t)/(t^2 +1))dt=∫_0 ^π ((te^(it) )/(t^2 +1))dt ⇒ ∫_0 ^π ((tsin t)/(t^2 +1))dt=imag (∫_0 ^π ((te^(it) )/(t^2 +1))dt) ...but I′m not good at this. Anyway we can get the exact result for ∫_(−∞) ^(+∞) ((tsin t)/(t^2 +1))dt=(π/e) using contour integration.

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

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

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

Commented by Frix last updated on 07/Aug/24

Yes, I also found this site but there′s no useable exact solution, we must approximate at the end.

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

Answered by Berbere last updated on 08/Aug/24

(t/(1+t^2 ))sin(t)=(1/2)((1/(t+i))+(1/(t−i)))sin(t) =Re∫_0 ^π ((sin(t))/(t+i))dt=∫_i ^(i+π) ((sin(u−i))/u)du =∫_i ^(i+π) ((sin(u)cos(i)−cos(u)sin(i))/u)du;cos(i)=ch(1) sin(i)=ish(1) =Re{∫_i ^(i+u) ch(1)((sinu)/u)du+ish(1)∫_i ^(i+u) ((−cos(u))/u)du} =Re{Si(i+π)−Si(i)}ch(1)+Re(ish(1)Ci(i)−Ci(i+π)}

$$\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{sin}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{t}+{i}}+\frac{\mathrm{1}}{{t}−{i}}\right){sin}\left({t}\right) \\ $$$$={Re}\overset{\pi} {\int}_{\mathrm{0}} \frac{{sin}\left({t}\right)}{{t}+{i}}{dt}=\int_{{i}} ^{{i}+\pi} \frac{{sin}\left({u}−{i}\right)}{{u}}{du} \\ $$$$=\int_{{i}} ^{{i}+\pi} \frac{{sin}\left({u}\right){cos}\left({i}\right)−{cos}\left({u}\right){sin}\left({i}\right)}{{u}}{du};{cos}\left({i}\right)={ch}\left(\mathrm{1}\right) \\ $$$${sin}\left({i}\right)={ish}\left(\mathrm{1}\right) \\ $$$$={Re}\left\{\int_{{i}} ^{{i}+{u}} {ch}\left(\mathrm{1}\right)\frac{{sinu}}{{u}}{du}+{ish}\left(\mathrm{1}\right)\int_{{i}} ^{{i}+{u}} \frac{−{cos}\left({u}\right)}{{u}}{du}\right\} \\ $$$$={Re}\left\{{Si}\left({i}+\pi\right)−{Si}\left({i}\right)\right\}{ch}\left(\mathrm{1}\right)+{Re}\left({ish}\left(\mathrm{1}\right){Ci}\left({i}\right)−{Ci}\left({i}+\pi\right)\right\} \\ $$

Commented by hardmath last updated on 11/Aug/24

Dear professor, your solutions are perfect

$$ \\ $$Dear professor, your solutions are perfect

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