Question Number 43906 by abdo.msup.com last updated on 17/Sep/18
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({x}\right)}{{sh}\left(\mathrm{2}{x}\right)}{dx} \\ $$
Commented by maxmathsup by imad last updated on 20/Sep/18
![let A = ∫_0 ^∞ ((sinx)/(sh(2x)))dx ⇒ A = ∫_0 ^∞ ((2sinx)/(e^(2x) −e^(−2x) )) dx = ∫_0 ^∞ ((e^(−2x) sinx)/(1−e^(−4x) ))dx =∫_0 ^∞ e^(−2x) (Σ_(n=0) ^∞ e^(−4nx) )sinx dx = Σ_(n=0) ^∞ ∫_0 ^∞ e^(−(4n+2)x) sinx dx =Σ_(n=0) ^∞ A_n and A_n =Im( ∫_0 ^∞ e^(−(4n+2 −i)x) dx) but ∫_0 ^∞ e^(−(4n+2 +i)x) dx =[−(1/(4n+2−i)) e^(−(4n+2+i)x) ]_0 ^(+∞) =(1/(4n+2−i)) ((4n+2+i)/((4n+2)^2 +1)) ⇒ A_n = (1/((4n+2)^2 +1)) ⇒ A = Σ_(n=0) ^∞ (1/(4(2n+1)^2 +1)) and this serie can be calculated by Fourier... be continued...](Q44061.png)
$${let}\:{A}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinx}}{{sh}\left(\mathrm{2}{x}\right)}{dx}\:\Rightarrow\:{A}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{sinx}}{{e}^{\mathrm{2}{x}} −{e}^{−\mathrm{2}{x}} }\:{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−\mathrm{2}{x}} \:{sinx}}{\mathrm{1}−{e}^{−\mathrm{4}{x}} }{dx}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\mathrm{2}{x}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−\mathrm{4}{nx}} \right){sinx}\:{dx} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\mathrm{4}{n}+\mathrm{2}\right){x}} {sinx}\:{dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \:{and}\:{A}_{{n}} ={Im}\left(\:\:\int_{\mathrm{0}} ^{\infty} {e}^{−\left(\mathrm{4}{n}+\mathrm{2}\:−{i}\right){x}} \:{dx}\right)\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\mathrm{4}{n}+\mathrm{2}\:+{i}\right){x}} {dx}\:=\left[−\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{2}−{i}}\:{e}^{−\left(\mathrm{4}{n}+\mathrm{2}+{i}\right){x}} \right]_{\mathrm{0}} ^{+\infty} =\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{2}−{i}} \\ $$$$\frac{\mathrm{4}{n}+\mathrm{2}+{i}}{\left(\mathrm{4}{n}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow\:{A}_{{n}} =\:\frac{\mathrm{1}}{\left(\mathrm{4}{n}+\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\:{A}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${and}\:{this}\:{serie}\:{can}\:{be}\:{calculated}\:\:{by}\:{Fourier}...\:{be}\:{continued}... \\ $$