nice-calculus-prove-that-cosh-px-cosh-x-pi-cos-pip-2-

Question Number 133334 by mnjuly1970 last updated on 21/Feb/21

... nice calculus... prove that:: 𝛗=∫_(−∞) ^( +∞) (( cosh(px))/(cosh(x))) = (π/(cos(((πp)/2))))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:\:\:\:\:{calculus}… \\ $$$$\:{prove}\:\:{that}:: \\ $$$$\:\:\:\boldsymbol{\phi}=\int_{−\infty} ^{\:+\infty} \frac{\:{cosh}\left({px}\right)}{{cosh}\left({x}\right)}\:=\:\frac{\pi}{{cos}\left(\frac{\pi{p}}{\mathrm{2}}\right)} \\ $$$$ \\ $$

Answered by mnjuly1970 last updated on 21/Feb/21

Commented by Dwaipayan Shikari last updated on 21/Feb/21

$${Haven}'{t}\:{thought}\:{of}\:.\:{Great}\:{way}\:{sir}! \\ $$

Commented by mnjuly1970 last updated on 21/Feb/21

$${sincerely}\:{yours}.. \\ $$$${grateful}\:{sir}\:{payan}… \\ $$

Answered by mathmax by abdo last updated on 22/Feb/21

Φ=∫_(−∞) ^(+∞) ((ch(px))/(ch(x)))dx ⇒Φ=∫_(−∞) ^(+∞) ((e^(px) +e^(−px) )/(e^x +e^(−x) ))dx =_(e^x =t) ∫_0 ^∞ ((t^p +t^(−p) )/(t+t^(−1) ))(dt/t) =∫_0 ^∞ ((t^p +t^(−p) )/(t^2 +1))dt =∫_0 ^∞ (t^p /(1+t^2 ))dt +∫_0 ^∞ (t^(−p) /(1+t^2 ))dt we have ∫_0 ^∞ (t^p /(1+t^2 ))dt =_(t=z^(1/2) ) (1/2)∫_0 ^∞ (z^(p/2) /(1+z))z^(−(1/2)) dz =(1/2)∫_0 ^∞ (z^(((p+1)/2)−1) /(1+z))dz =(1/2)(π/(sin(π(((p+1)/2)))))=(π/(2cos(((pπ)/2)))) also ∫_0 ^∞ (t^(−p) /(1+t^2 ))dt =(π/(2cos(((−pπ)/2)))) =(π/(2cos(((pπ)/2)))) ⇒ Φ=(π/(2cos(((pπ)/2))))+(π/(2cos(((pπ)/2)))) =(π/(cos(((pπ)/2))))

$$\Phi=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{ch}\left(\mathrm{px}\right)}{\mathrm{ch}\left(\mathrm{x}\right)}\mathrm{dx}\:\Rightarrow\Phi=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{px}} \:+\mathrm{e}^{−\mathrm{px}} }{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }\mathrm{dx} \\ $$$$=_{\mathrm{e}^{\mathrm{x}} \:=\mathrm{t}} \:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{p}} \:+\mathrm{t}^{−\mathrm{p}} }{\mathrm{t}+\mathrm{t}^{−\mathrm{1}} }\frac{\mathrm{dt}}{\mathrm{t}}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{p}} \:+\mathrm{t}^{−\mathrm{p}} }{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{p}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:+\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{−\mathrm{p}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\mathrm{we}\:\mathrm{have}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{p}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=_{\mathrm{t}=\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{2}}} } \:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{z}^{\frac{\mathrm{p}}{\mathrm{2}}} }{\mathrm{1}+\mathrm{z}}\mathrm{z}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{z}^{\frac{\mathrm{p}+\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+\mathrm{z}}\mathrm{dz}\:=\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi}{\mathrm{sin}\left(\pi\left(\frac{\mathrm{p}+\mathrm{1}}{\mathrm{2}}\right)\right)}=\frac{\pi}{\mathrm{2cos}\left(\frac{\mathrm{p}\pi}{\mathrm{2}}\right)}\:\mathrm{also} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{−\mathrm{p}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=\frac{\pi}{\mathrm{2cos}\left(\frac{−\mathrm{p}\pi}{\mathrm{2}}\right)}\:=\frac{\pi}{\mathrm{2cos}\left(\frac{\mathrm{p}\pi}{\mathrm{2}}\right)}\:\Rightarrow \\ $$$$\Phi=\frac{\pi}{\mathrm{2cos}\left(\frac{\mathrm{p}\pi}{\mathrm{2}}\right)}+\frac{\pi}{\mathrm{2cos}\left(\frac{\mathrm{p}\pi}{\mathrm{2}}\right)}\:=\frac{\pi}{\mathrm{cos}\left(\frac{\mathrm{p}\pi}{\mathrm{2}}\right)} \\ $$$$ \\ $$

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