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

                ... nice      calculus...   prove  that::     𝛗=∫_(βˆ’βˆž) ^( +∞) (( cosh(px))/(cosh(x))) = (Ο€/(cos(((Ο€p)/2))))

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

Answered by mnjuly1970 last updated on 21/Feb/21

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

Havenβ€²t thought of . Great way sir!

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

Commented by mnjuly1970 last updated on 21/Feb/21

sincerely yours..  grateful sir payan...

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

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

Ξ¦=∫_(βˆ’βˆž) ^(+∞)  ((ch(px))/(ch(x)))dx β‡’Ξ¦=∫_(βˆ’βˆž) ^(+∞)  ((e^(px)  +e^(βˆ’px) )/(e^x +e^(βˆ’x) ))dx  =_(e^x  =t)     ∫_0 ^∞  ((t^p  +t^(βˆ’p) )/(t+t^(βˆ’1) ))(dt/t) =∫_0 ^∞  ((t^p  +t^(βˆ’p) )/(t^2  +1))dt =∫_0 ^∞  (t^p /(1+t^2 ))dt +∫_0 ^∞  (t^(βˆ’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^(βˆ’(1/2)) dz  =(1/2)∫_0 ^∞  (z^(((p+1)/2)βˆ’1) /(1+z))dz =(1/2)(Ο€/(sin(Ο€(((p+1)/2)))))=(Ο€/(2cos(((pΟ€)/2)))) also  ∫_0 ^∞ (t^(βˆ’p) /(1+t^2 ))dt =(Ο€/(2cos(((βˆ’pΟ€)/2)))) =(Ο€/(2cos(((pΟ€)/2)))) β‡’  Ξ¦=(Ο€/(2cos(((pΟ€)/2))))+(Ο€/(2cos(((pΟ€)/2)))) =(Ο€/(cos(((pΟ€)/2))))

$$\Phi=\int_{βˆ’\infty} ^{+\infty} \:\frac{\mathrm{ch}\left(\mathrm{px}\right)}{\mathrm{ch}\left(\mathrm{x}\right)}\mathrm{dx}\:\Rightarrow\Phi=\int_{βˆ’\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{px}} \:+\mathrm{e}^{βˆ’\mathrm{px}} }{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{βˆ’\mathrm{x}} }\mathrm{dx} \\ $$$$=_{\mathrm{e}^{\mathrm{x}} \:=\mathrm{t}} \:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{p}} \:+\mathrm{t}^{βˆ’\mathrm{p}} }{\mathrm{t}+\mathrm{t}^{βˆ’\mathrm{1}} }\frac{\mathrm{dt}}{\mathrm{t}}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{p}} \:+\mathrm{t}^{βˆ’\mathrm{p}} }{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{p}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:+\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{βˆ’\mathrm{p}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\mathrm{we}\:\mathrm{have}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{p}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=_{\mathrm{t}=\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{2}}} } \:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{z}^{\frac{\mathrm{p}}{\mathrm{2}}} }{\mathrm{1}+\mathrm{z}}\mathrm{z}^{βˆ’\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{z}^{\frac{\mathrm{p}+\mathrm{1}}{\mathrm{2}}βˆ’\mathrm{1}} }{\mathrm{1}+\mathrm{z}}\mathrm{dz}\:=\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi}{\mathrm{sin}\left(\pi\left(\frac{\mathrm{p}+\mathrm{1}}{\mathrm{2}}\right)\right)}=\frac{\pi}{\mathrm{2cos}\left(\frac{\mathrm{p}\pi}{\mathrm{2}}\right)}\:\mathrm{also} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{βˆ’\mathrm{p}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=\frac{\pi}{\mathrm{2cos}\left(\frac{βˆ’\mathrm{p}\pi}{\mathrm{2}}\right)}\:=\frac{\pi}{\mathrm{2cos}\left(\frac{\mathrm{p}\pi}{\mathrm{2}}\right)}\:\Rightarrow \\ $$$$\Phi=\frac{\pi}{\mathrm{2cos}\left(\frac{\mathrm{p}\pi}{\mathrm{2}}\right)}+\frac{\pi}{\mathrm{2cos}\left(\frac{\mathrm{p}\pi}{\mathrm{2}}\right)}\:=\frac{\pi}{\mathrm{cos}\left(\frac{\mathrm{p}\pi}{\mathrm{2}}\right)} \\ $$$$ \\ $$

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