Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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...

$${thank}\:{you}\:{so}\:{much} \\ $$$$\:{i}\:{will}\:{rechck}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com