Question Number 152797 by mnjuly1970 last updated on 01/Sep/21
$$ \\ $$$$\:\:\:{nice}..{mathematics}… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{Prove}\:\mathrm{that}… \\ $$$$\:\mathrm{I}=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{cos}\:\left({x}\:\right)}{{cosh}\:\left({x}\:\right)}\:{dx}=\frac{\pi}{\:{cosh}\:\left(\frac{\pi}{\mathrm{2}}\:\right)}\:…….\blacksquare\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:{prepared}\:::\:\:{m}.{n} \\ $$$$ \\ $$
Answered by mnjuly1970 last updated on 01/Sep/21
![I:= ∫_0 ^( ∞) (( e^( ix) +e^( −ix) )/(e^( x) +e^( −x) ))dx= ∫_0 ^( ∞) (( e^( x(i−1)) +e^( −x(i+1)) )/(1+e^( −2x) ))dx :=^(e^( −2x) =y) (1/2)∫_0 ^( 1) (( y^( ((1−i)/2)) +y^((1+i)/2) )/(1+ y)) (dy/y) := (1/2) ∫_0 ^( 1) (( y^( −((1+i)/2)) +y^( −((1−i)/2)) )/(1+y)) dy := (1/2) ∫_0 ^( 1) (( (1−y )( y^( −((1+i)/2)) +y^(−((1−i)/2)) ))/(1−y^( 2) ))dy := (1/2) ∫_0 ^( 1) (( y^( −((1+i)/2)) +y^(−((1−i)/2)) −y^((1−i)/2) −y^( ((1+i)/2)) )/(1−y^( 2) ))dy :=^(y^( 2) = t) (1/4) ∫_0 ^( 1) (( t^( ((−3−i)/4)) −1+t^((−3+i)/4) −1+1−t^((−1−i)/4) +1−t^( ((−1+i)/4)) )/(1−t))dt := (1/4) (ψ (((3−i)/4) )+[ψ(((3+i)/4))−ψ(((1−i)/4))]−ψ(((1+i)/4))) :=^(ψ (z )−ψ (1−z )= −π cot(πz )) (1/4) (−πcot ((((3+i)π)/4))−πcot((((3−i)π)/4) )) :=(π/4) {cot((π/4) −((iπ)/4) )+ cot((π/4) +((iπ)/4))} := (π/4) {(( 1+tan(((iπ)/4)))/(1−tan(((iπ)/4) ))) +((1−tan(((iπ)/4)))/(1+tan(((iπ)/4)))) } := (π/4) { (( 2( 1+tan^( 2) (((iπ)/4) )))/(1−tan^( 2) (((iπ)/4))))} := (π/2) ((( 1)/(cos (i(π/2)))))= (π/(2cosh ((π/2) ))) ...■ m.n](https://www.tinkutara.com/question/Q152813.png)
$$\:\:\:\:\:\:\:\:\:\mathrm{I}:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{e}^{\:\:{ix}} +{e}^{\:−{ix}} }{{e}^{\:{x}} +{e}^{\:−{x}} }{dx}=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{e}^{\:{x}\left({i}−\mathrm{1}\right)} +{e}^{\:−{x}\left({i}+\mathrm{1}\right)} }{\mathrm{1}+{e}^{\:−\mathrm{2}{x}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\::\overset{{e}^{\:−\mathrm{2}{x}} ={y}} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{y}^{\:\frac{\mathrm{1}−{i}}{\mathrm{2}}} +{y}^{\frac{\mathrm{1}+{i}}{\mathrm{2}}} }{\mathrm{1}+\:{y}}\:\frac{{dy}}{{y}} \\ $$$$\:\:\:\:\:\::=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{y}^{\:−\frac{\mathrm{1}+{i}}{\mathrm{2}}} +{y}^{\:−\frac{\mathrm{1}−{i}}{\mathrm{2}}} }{\mathrm{1}+{y}}\:{dy} \\ $$$$\:\:\:\:\:\::=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\left(\mathrm{1}−{y}\:\right)\left(\:{y}^{\:−\frac{\mathrm{1}+{i}}{\mathrm{2}}} +{y}^{−\frac{\mathrm{1}−{i}}{\mathrm{2}}} \right)}{\mathrm{1}−{y}^{\:\mathrm{2}} }{dy} \\ $$$$\:\:\:\::=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{y}^{\:−\frac{\mathrm{1}+{i}}{\mathrm{2}}} +{y}^{−\frac{\mathrm{1}−{i}}{\mathrm{2}}} −{y}^{\frac{\mathrm{1}−{i}}{\mathrm{2}}} −{y}^{\:\frac{\mathrm{1}+{i}}{\mathrm{2}}} }{\mathrm{1}−{y}^{\:\mathrm{2}} }{dy} \\ $$$$\:\:\:\::\overset{{y}^{\:\mathrm{2}} =\:{t}} {=}\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{t}^{\:\frac{−\mathrm{3}−{i}}{\mathrm{4}}} −\mathrm{1}+{t}^{\frac{−\mathrm{3}+{i}}{\mathrm{4}}} −\mathrm{1}+\mathrm{1}−{t}^{\frac{−\mathrm{1}−{i}}{\mathrm{4}}} +\mathrm{1}−{t}^{\:\frac{−\mathrm{1}+{i}}{\mathrm{4}}} }{\mathrm{1}−{t}}{dt} \\ $$$$\:\:\:\:\:\::=\:\frac{\mathrm{1}}{\mathrm{4}}\:\left(\psi\:\left(\frac{\mathrm{3}−{i}}{\mathrm{4}}\:\right)+\left[\psi\left(\frac{\mathrm{3}+{i}}{\mathrm{4}}\right)−\psi\left(\frac{\mathrm{1}−{i}}{\mathrm{4}}\right)\right]−\psi\left(\frac{\mathrm{1}+{i}}{\mathrm{4}}\right)\right) \\ $$$$\:\:\:\::\overset{\psi\:\left({z}\:\right)−\psi\:\left(\mathrm{1}−{z}\:\right)=\:−\pi\:{cot}\left(\pi{z}\:\right)} {=}\:\frac{\mathrm{1}}{\mathrm{4}}\:\left(−\pi{cot}\:\left(\frac{\left(\mathrm{3}+{i}\right)\pi}{\mathrm{4}}\right)−\pi{cot}\left(\frac{\left(\mathrm{3}−{i}\right)\pi}{\mathrm{4}}\:\right)\right) \\ $$$$\:\:\:\:\::=\frac{\pi}{\mathrm{4}}\:\left\{{cot}\left(\frac{\pi}{\mathrm{4}}\:−\frac{{i}\pi}{\mathrm{4}}\:\right)+\:{cot}\left(\frac{\pi}{\mathrm{4}}\:+\frac{{i}\pi}{\mathrm{4}}\right)\right\} \\ $$$$\:\:\:\:\::=\:\frac{\pi}{\mathrm{4}}\:\left\{\frac{\:\mathrm{1}+{tan}\left(\frac{{i}\pi}{\mathrm{4}}\right)}{\mathrm{1}−{tan}\left(\frac{{i}\pi}{\mathrm{4}}\:\right)}\:+\frac{\mathrm{1}−{tan}\left(\frac{{i}\pi}{\mathrm{4}}\right)}{\mathrm{1}+{tan}\left(\frac{{i}\pi}{\mathrm{4}}\right)}\:\right\} \\ $$$$\:\:\:\:\:\::=\:\frac{\pi}{\mathrm{4}}\:\left\{\:\frac{\:\mathrm{2}\left(\:\mathrm{1}+{tan}^{\:\mathrm{2}} \left(\frac{{i}\pi}{\mathrm{4}}\:\right)\right)}{\mathrm{1}−{tan}^{\:\mathrm{2}} \left(\frac{{i}\pi}{\mathrm{4}}\right)}\right\} \\ $$$$\:\:\:\:\:\:\:\::=\:\frac{\pi}{\mathrm{2}}\:\left(\frac{\:\mathrm{1}}{{cos}\:\left({i}\frac{\pi}{\mathrm{2}}\right)}\right)=\:\frac{\pi}{\mathrm{2}{cosh}\:\left(\frac{\pi}{\mathrm{2}}\:\right)}\:…\blacksquare\:{m}.{n} \\ $$