......nice .... calculus...... prove that:: ξ := ∫_(−∞) ^( ∞) ((cos (πx^2 ))/(cosh(πx)))dx=(1/( (√2))) .... .......


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Question Number 140831 by mnjuly1970 last updated on 13/May/21

$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:......{nice}\:....\:{calculus}...... \\ $$$$\:\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:\:\:\:\:\xi\::=\:\int_{−\infty} ^{\:\infty} \frac{{cos}\:\left(\pi{x}^{\mathrm{2}} \right)}{{cosh}\left(\pi{x}\right)}{dx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:.... \\ $$$$\:\:\:\:....... \\ $$
	Answered by mathmax by abdo last updated on 13/May/21
	$Φ=∫_(−∞) ^(+∞) ((cos(πx^2 ))/(ch(πx)))dx ⇒Φ=_(πx=t) 2∫_0 ^(+∞) ((cos(t.(t/π)))/(ch(t)))(dt/π) =(2/π) ∫_0 ^(+∞) ((cos((t^2 /π)))/(ch(t)))dt ⇒ Φ=(4/π)∫_0 ^(+∞) ((cos((t^2 /π)))/(e^t +e^(−t) ))dt =(4/π) ∫_0 ^∞ ((e^(−t) cos((t^2 /π)))/(1+e^(−2t) )) dt ⇒(π/4)Φ=∫_0 ^∞ e^(−t) cos((t^2 /π))Σ_(n=0) ^∞ e^(−2nt) dt =Σ_(n=0) ^∞ ∫_0 ^∞ e^(−(2n+1)t) cos((t^2 /π))dt =_((2n+1)t=y) Σ_(n=0) ^∞ ∫_0 ^∞ e^(−y) cos((y^2 /(π(2n+1)^2 )))(dy/((2n+1))) =Σ_(n=0) ^∞ (1/(2n+1)) Re(∫_0 ^∞ e^(−y−i(y^2 /(π(2n+1)^2 ))) dy) we have ∫_0 ^∞ e^(−i(y^2 /(π(2n+1)^2 ))−y) dy =∫_0 ^∞ e^(−i(((y/( (√π)(2n+1))))^2 −iy)) dy =∫_0 ^∞ e^(−i{ ((y/( (√π)(2n+1))))^2 −2((y/( (√π)(2n+1))))i(√π)(2n+1)+π(2n+1)^2 −π(2n+1)^2 }) dy =∫_0 ^∞ e^(−i{((y/( (√π)(2n+1)))−(√π)(2n+1))^2 −π(2n+1)^2 }) dy =e^(iπ(2n+1)^(2 ) ) ∫_0 ^∞ e^(−i{(y/( (√π)(2n+1)))−(√π)(2n+1)}^2 ) dy =_((y/( (√π)(2n+1)))−(√π)(2n+1) =z) e^(iπ(2n+)^2 ) ∫_(−(√π)(2n+1)) ^(+∞) e^(−iz^2 ) (√π)(2n+1)dz ....be continued...$
	$$\Phi=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\pi\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{ch}\left(\pi\mathrm{x}\right)}\mathrm{dx}\:\Rightarrow\Phi=_{\pi\mathrm{x}=\mathrm{t}} \:\:\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{t}.\frac{\mathrm{t}}{\pi}\right)}{\mathrm{ch}\left(\mathrm{t}\right)}\frac{\mathrm{dt}}{\pi} \\ $$$$=\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{+\infty} \:\frac{\mathrm{cos}\left(\frac{\mathrm{t}^{\mathrm{2}} }{\pi}\right)}{\mathrm{ch}\left(\mathrm{t}\right)}\mathrm{dt}\:\Rightarrow\:\Phi=\frac{\mathrm{4}}{\pi}\int_{\mathrm{0}} ^{+\infty} \:\frac{\mathrm{cos}\left(\frac{\mathrm{t}^{\mathrm{2}} }{\pi}\right)}{\mathrm{e}^{\mathrm{t}} \:+\mathrm{e}^{−\mathrm{t}} }\mathrm{dt} \\ $$$$=\frac{\mathrm{4}}{\pi}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{e}^{−\mathrm{t}} \:\mathrm{cos}\left(\frac{\mathrm{t}^{\mathrm{2}} }{\pi}\right)}{\mathrm{1}+\mathrm{e}^{−\mathrm{2t}} }\:\mathrm{dt}\:\Rightarrow\frac{\pi}{\mathrm{4}}\Phi=\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{t}} \:\mathrm{cos}\left(\frac{\mathrm{t}^{\mathrm{2}} }{\pi}\right)\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{2nt}} \:\mathrm{dt} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{t}} \:\mathrm{cos}\left(\frac{\mathrm{t}^{\mathrm{2}} }{\pi}\right)\mathrm{dt} \\ $$$$=_{\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{t}=\mathrm{y}} \:\:\:\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{y}} \:\mathrm{cos}\left(\frac{\mathrm{y}^{\mathrm{2}} }{\pi\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\frac{\mathrm{dy}}{\left(\mathrm{2n}+\mathrm{1}\right)} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\:\mathrm{Re}\left(\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{y}−\mathrm{i}\frac{\mathrm{y}^{\mathrm{2}} }{\pi\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }} \mathrm{dy}\right)\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\mathrm{i}\frac{\mathrm{y}^{\mathrm{2}} }{\pi\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{y}} \mathrm{dy}\:=\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\mathrm{i}\left(\left(\frac{\mathrm{y}}{\:\sqrt{\pi}\left(\mathrm{2n}+\mathrm{1}\right)}\right)^{\mathrm{2}} −\mathrm{iy}\right)} \mathrm{dy} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{i}\left\{\:\left(\frac{\mathrm{y}}{\:\sqrt{\pi}\left(\mathrm{2n}+\mathrm{1}\right)}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{y}}{\:\sqrt{\pi}\left(\mathrm{2n}+\mathrm{1}\right)}\right)\mathrm{i}\sqrt{\pi}\left(\mathrm{2n}+\mathrm{1}\right)+\pi\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} −\pi\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} \right\}} \mathrm{dy} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\mathrm{i}\left\{\left(\frac{\mathrm{y}}{\:\sqrt{\pi}\left(\mathrm{2n}+\mathrm{1}\right)}−\sqrt{\pi}\left(\mathrm{2n}+\mathrm{1}\right)\right)^{\mathrm{2}} −\pi\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} \right\}} \mathrm{dy} \\ $$$$=\mathrm{e}^{\mathrm{i}\pi\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}\:} \:} \int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{i}\left\{\frac{\mathrm{y}}{\:\sqrt{\pi}\left(\mathrm{2n}+\mathrm{1}\right)}−\sqrt{\pi}\left(\mathrm{2n}+\mathrm{1}\right)\right\}^{\mathrm{2}} } \mathrm{dy} \\ $$$$=_{\frac{\mathrm{y}}{\:\sqrt{\pi}\left(\mathrm{2n}+\mathrm{1}\right)}−\sqrt{\pi}\left(\mathrm{2n}+\mathrm{1}\right)\:=\mathrm{z}} \mathrm{e}^{\mathrm{i}\pi\left(\mathrm{2n}+\right)^{\mathrm{2}} } \:\:\:\int_{−\sqrt{\pi}\left(\mathrm{2n}+\mathrm{1}\right)} ^{+\infty} \:\mathrm{e}^{−\mathrm{iz}^{\mathrm{2}} } \sqrt{\pi}\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{dz}\: \\ $$$$....\mathrm{be}\:\mathrm{continued}... \\ $$
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