nice-calculus-prove-that-0-cos-2x-cosh-x-dx-pi-2cosh-pi-

Question Number 132023 by mnjuly1970 last updated on 10/Feb/21

.....nice calculus..... prove that::: 𝛗=∫_0 ^( ∞) ((cos(2x))/(cosh(x))) dx=^? (π/(2cosh(π)))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..{nice}\:\:{calculus}….. \\ $$$${prove}\:{that}::: \\ $$$$\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left(\mathrm{2}{x}\right)}{{cosh}\left({x}\right)}\:{dx}\overset{?} {=}\frac{\pi}{\mathrm{2}{cosh}\left(\pi\right)} \\ $$$$ \\ $$

Answered by mindispower last updated on 10/Feb/21

(1/(ch(x)))=((2e^x )/(e^(2x) +1))=2e^(−x) Σ_(k≥0) (−1)^k e^(−2kx) =Σ_(k≥0) (−1)^k e^(−x(2k+1)) ....true ∀x>0 φ=(1/2)Re∫_(−∞) ^∞ 2((e^(2ix+x) )/(e^(2x) +1))dx let C ={x,Imx≥0} e^(2x) +1=0⇒2x=iπ+2ikπ x_k =i((π/2)+kπ),k≥0 Res((e^(2ix+x) /(e^(2x) +1)),x_k )=((e^(−π−2kπ) .i.(−1)^k )/(2(−1))) ∫_C (e^(2ix+x) /(e^(2x) +1))dx=2iπRes((e^(2ix+x) /(e^(2x) +1)),x∈C) =2iπΣ_(k≥0) ((e^(−π−2kπ) i(−1)^k )/(−2)) =πe^(−π) Σ_(k≥0) (−e^(−2π) )^k =((πe^(−π) )/(1+e^(−2π) ))=(π/(e^π +e^(−π) ))=(π/(2ch(π))) we get so ∫_0 ^∞ ((cos(2x))/(ch(x)))dx=(1/2).Re.2(π/(2ch(π)))=(π/(2cosh(π)))

$$\frac{\mathrm{1}}{{ch}\left({x}\right)}=\frac{\mathrm{2}{e}^{{x}} }{{e}^{\mathrm{2}{x}} +\mathrm{1}}=\mathrm{2}{e}^{−{x}} \underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} {e}^{−\mathrm{2}{kx}} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} {e}^{−{x}\left(\mathrm{2}{k}+\mathrm{1}\right)} ….{true}\:\forall{x}>\mathrm{0} \\ $$$$\phi=\frac{\mathrm{1}}{\mathrm{2}}{Re}\int_{−\infty} ^{\infty} \mathrm{2}\frac{{e}^{\mathrm{2}{ix}+{x}} \:}{{e}^{\mathrm{2}{x}} +\mathrm{1}}{dx} \\ $$$${let}\:{C}\:=\left\{{x},{Imx}\geqslant\mathrm{0}\right\} \\ $$$${e}^{\mathrm{2}{x}} +\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{2}{x}={i}\pi+\mathrm{2}{ik}\pi \\ $$$${x}_{{k}} ={i}\left(\frac{\pi}{\mathrm{2}}+{k}\pi\right),{k}\geqslant\mathrm{0} \\ $$$${Res}\left(\frac{{e}^{\mathrm{2}{ix}+{x}} }{{e}^{\mathrm{2}{x}} +\mathrm{1}},{x}_{{k}} \right)=\frac{{e}^{−\pi−\mathrm{2}{k}\pi} .{i}.\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}\left(−\mathrm{1}\right)} \\ $$$$\int_{{C}} \frac{{e}^{\mathrm{2}{ix}+{x}} }{{e}^{\mathrm{2}{x}} +\mathrm{1}}{dx}=\mathrm{2}{i}\pi{Res}\left(\frac{{e}^{\mathrm{2}{ix}+{x}} }{{e}^{\mathrm{2}{x}} +\mathrm{1}},{x}\in{C}\right) \\ $$$$=\mathrm{2}{i}\pi\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{e}^{−\pi−\mathrm{2}{k}\pi} {i}\left(−\mathrm{1}\right)^{{k}} }{−\mathrm{2}} \\ $$$$=\pi{e}^{−\pi} \underset{{k}\geqslant\mathrm{0}} {\sum}\left(−{e}^{−\mathrm{2}\pi} \right)^{{k}} =\frac{\pi{e}^{−\pi} }{\mathrm{1}+{e}^{−\mathrm{2}\pi} }=\frac{\pi}{{e}^{\pi} +{e}^{−\pi} }=\frac{\pi}{\mathrm{2}{ch}\left(\pi\right)} \\ $$$${we}\:{get}\:{so} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left(\mathrm{2}{x}\right)}{{ch}\left({x}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{2}}.{Re}.\mathrm{2}\frac{\pi}{\mathrm{2}{ch}\left(\pi\right)}=\frac{\pi}{\mathrm{2}{cosh}\left(\pi\right)} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 10/Feb/21

$${excellent}.. \\ $$$${thanks}\:{alot}\:{sir}\:{power}… \\ $$

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