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Question Number 141649 by mnjuly1970 last updated on 21/May/21
.......advanced calculus...... prove that−:: φ:=∫_0 ^( ∞) ((cos(2πx^2 ))/(cosh^2 (πx)))dx=(1/4) ✓
$$\:\:\:\:\:\:\:\:\:…….{advanced}\:\:{calculus}…… \\ $$$$\:\:\:\:{prove}\:\:{that}−:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\phi:=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{cos}\left(\mathrm{2}\pi{x}^{\mathrm{2}} \right)}{{cosh}^{\mathrm{2}} \left(\pi{x}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\checkmark \\ $$
Answered by ArielVyny last updated on 22/May/21
φ=∫_0 ^∝ ((cos(2πx^2 )+1)/(cosh^2 (πx)))dx−∫_0 ^∝ (1/(cosh^2 (πx)))dx φ=φ_1 −φ_2 φ_2 =∫_0 ^∝ (1/(cosh^2 (πx)))dx πx=t πdx=dt φ_2 =(1/π)∫_0 ^∝ (1/(cosh^2 (t)))dt=(1/π) φ_1 =∫_0 ^∝ ((cos(2πx^2 )+1)/(cosh^2 (πx)))dx cos^2 (πx^2 )=((1+cos(2πx^2 ))/2) φ_1 =∫_0 ^∝ ((2cos^2 (πx^2 ))/(cosh^2 (πx)))dx φ_1 =2∫_0 ^∝ (((cos(πx^2 ))/(cosh(πx))))^2 dx φ_1 =2∫_0 ^∝ ((((−1)^x^2 )/(cosh(πx))))^2 dx φ_1 =2∫_0 ^∝ ((1/(cosh(πx))))^2 dx φ_1 =2∫_0 ^∝ (1/(cosh^2 (πx))dx πx=t. πdx=dt φ_1 =2(1/π)∫_0 ^∝ (1/(cosh^2 (t)))dt φ_1 =(2/π) φ=(1/π) check in the answer because i do not have your result.
$$ \\ $$$$\phi=\int_{\mathrm{0}} ^{\propto} \frac{{cos}\left(\mathrm{2}\pi{x}^{\mathrm{2}} \right)+\mathrm{1}}{{cosh}^{\mathrm{2}} \left(\pi{x}\right)}{dx}−\int_{\mathrm{0}} ^{\propto} \frac{\mathrm{1}}{{cosh}^{\mathrm{2}} \left(\pi{x}\right)}{dx} \\ $$$$\phi=\phi_{\mathrm{1}} −\phi_{\mathrm{2}} \\ $$$$\phi_{\mathrm{2}} =\int_{\mathrm{0}} ^{\propto} \frac{\mathrm{1}}{{cosh}^{\mathrm{2}} \left(\pi{x}\right)}{dx}\:\: \\ $$$$\pi{x}={t}\:\:\pi{dx}={dt} \\ $$$$\phi_{\mathrm{2}} =\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\propto} \frac{\mathrm{1}}{{cosh}^{\mathrm{2}} \left({t}\right)}{dt}=\frac{\mathrm{1}}{\pi} \\ $$$$\phi_{\mathrm{1}} =\int_{\mathrm{0}} ^{\propto} \frac{{cos}\left(\mathrm{2}\pi{x}^{\mathrm{2}} \right)+\mathrm{1}}{{cosh}^{\mathrm{2}} \left(\pi{x}\right)}{dx} \\ $$$${cos}^{\mathrm{2}} \left(\pi{x}^{\mathrm{2}} \right)=\frac{\mathrm{1}+{cos}\left(\mathrm{2}\pi{x}^{\mathrm{2}} \right)}{\mathrm{2}} \\ $$$$\phi_{\mathrm{1}} =\int_{\mathrm{0}} ^{\propto} \frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\pi{x}^{\mathrm{2}} \right)}{{cosh}^{\mathrm{2}} \left(\pi{x}\right)}{dx} \\ $$$$\phi_{\mathrm{1}} =\mathrm{2}\int_{\mathrm{0}} ^{\propto} \left(\frac{{cos}\left(\pi{x}^{\mathrm{2}} \right)}{{cosh}\left(\pi{x}\right)}\right)^{\mathrm{2}} {dx} \\ $$$$\phi_{\mathrm{1}} =\mathrm{2}\int_{\mathrm{0}} ^{\propto} \left(\frac{\left(−\mathrm{1}\right)^{{x}^{\mathrm{2}} } }{{cosh}\left(\pi{x}\right)}\right)^{\mathrm{2}} {dx} \\ $$$$\phi_{\mathrm{1}} =\mathrm{2}\int_{\mathrm{0}} ^{\propto} \left(\frac{\mathrm{1}}{{cosh}\left(\pi{x}\right)}\right)^{\mathrm{2}} {dx} \\ $$$$\phi_{\mathrm{1}} =\mathrm{2}\int_{\mathrm{0}} ^{\propto} \frac{\mathrm{1}}{{cosh}^{\mathrm{2}} \left(\pi{x}\right.}{dx} \\ $$$$\pi{x}={t}.\:\:\pi{dx}={dt} \\ $$$$\phi_{\mathrm{1}} =\mathrm{2}\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\propto} \frac{\mathrm{1}}{{cosh}^{\mathrm{2}} \left({t}\right)}{dt} \\ $$$$\phi_{\mathrm{1}} =\frac{\mathrm{2}}{\pi} \\ $$$$\phi=\frac{\mathrm{1}}{\pi} \\ $$$${check}\:{in}\:{the}\:{answer}\:{because}\:{i}\:{do}\:{not}\:{have}\:{your}\:{result}. \\ $$
Commented by mnjuly1970 last updated on 22/May/21
thank you so much i will rechck...
$${thank}\:{you}\:{so}\:{much} \\ $$$$\:{i}\:{will}\:{rechck}… \\ $$

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