......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

$. . . . . . n i c e . . . . c a l c u l u s . . . . . .$ $p r o v e t h a t ::$ $ξ := \int_{- \infty}^{\infty} \frac{c o s (π x^{2})}{c o s h (π x)} d x = \frac{1}{\sqrt{2}} . . . .$ $. . . . . . .$
	Answered by mathmax by abdo last updated on 13/May/21
	$Φ=∫_(−∞) ^(+∞) ((cos(πx^2 ))/(ch(πx)))dx ⇒Φ=_(πx=t) 2∫_0 ^(+∞) ((cos(t.(t/π)))/(ch(t)))(dt/π) =(2/π) ∫_0 ^(+∞) ((cos((t^2 /π)))/(ch(t)))dt ⇒ Φ=(4/π)∫_0 ^(+∞) ((cos((t^2 /π)))/(e^t +e^(−t) ))dt =(4/π) ∫_0 ^∞ ((e^(−t) cos((t^2 /π)))/(1+e^(−2t) )) dt ⇒(π/4)Φ=∫_0 ^∞ e^(−t) cos((t^2 /π))Σ_(n=0) ^∞ e^(−2nt) dt =Σ_(n=0) ^∞ ∫_0 ^∞ e^(−(2n+1)t) cos((t^2 /π))dt =_((2n+1)t=y) Σ_(n=0) ^∞ ∫_0 ^∞ e^(−y) cos((y^2 /(π(2n+1)^2 )))(dy/((2n+1))) =Σ_(n=0) ^∞ (1/(2n+1)) Re(∫_0 ^∞ e^(−y−i(y^2 /(π(2n+1)^2 ))) dy) we have ∫_0 ^∞ e^(−i(y^2 /(π(2n+1)^2 ))−y) dy =∫_0 ^∞ e^(−i(((y/( (√π)(2n+1))))^2 −iy)) dy =∫_0 ^∞ e^(−i{ ((y/( (√π)(2n+1))))^2 −2((y/( (√π)(2n+1))))i(√π)(2n+1)+π(2n+1)^2 −π(2n+1)^2 }) dy =∫_0 ^∞ e^(−i{((y/( (√π)(2n+1)))−(√π)(2n+1))^2 −π(2n+1)^2 }) dy =e^(iπ(2n+1)^(2 ) ) ∫_0 ^∞ e^(−i{(y/( (√π)(2n+1)))−(√π)(2n+1)}^2 ) dy =_((y/( (√π)(2n+1)))−(√π)(2n+1) =z) e^(iπ(2n+)^2 ) ∫_(−(√π)(2n+1)) ^(+∞) e^(−iz^2 ) (√π)(2n+1)dz ....be continued...$
	$Φ = \int_{- \infty}^{+ \infty} \frac{\cos (π x^{2})}{ch (π x)} dx \Rightarrow Φ =_{π x = t} 2 \int_{0}^{+ \infty} \frac{\cos (t . \frac{t}{π})}{ch (t)} \frac{dt}{π}$ $= \frac{2}{π} \int_{0}^{+ \infty} \frac{\cos (\frac{t^{2}}{π})}{ch (t)} dt \Rightarrow Φ = \frac{4}{π} \int_{0}^{+ \infty} \frac{\cos (\frac{t^{2}}{π})}{e^{t} + e^{- t}} dt$ $= \frac{4}{π} \int_{0}^{\infty} \frac{e^{- t} \cos (\frac{t^{2}}{π})}{1 + e^{- 2 t}} dt \Rightarrow \frac{π}{4} Φ = \int_{0}^{\infty} e^{- t} \cos (\frac{t^{2}}{π}) \sum_{n = 0}^{\infty} e^{- 2 nt} dt$ $= \sum_{n = 0}^{\infty} \int_{0}^{\infty} e^{- (2 n + 1) t} \cos (\frac{t^{2}}{π}) dt$ $=_{(2 n + 1) t = y} \sum_{n = 0}^{\infty} \int_{0}^{\infty} e^{- y} \cos (\frac{y^{2}}{π {(2 n + 1)}^{2}}) \frac{dy}{(2 n + 1)}$ $= \sum_{n = 0}^{\infty} \frac{1}{2 n + 1} Re (\int_{0}^{\infty} e^{- y - i \frac{y^{2}}{π {(2 n + 1)}^{2}}} dy) we have$ $\int_{0}^{\infty} e^{- i \frac{y^{2}}{π {(2 n + 1)}^{2}} - y} dy = \int_{0}^{\infty} e^{- i ({(\frac{y}{\sqrt{π} (2 n + 1)})}^{2} - iy)} dy$ $= \int_{0}^{\infty} e^{- i {{(\frac{y}{\sqrt{π} (2 n + 1)})}^{2} - 2 (\frac{y}{\sqrt{π} (2 n + 1)}) i \sqrt{π} (2 n + 1) + π {(2 n + 1)}^{2} - π {(2 n + 1)}^{2}}} dy$ $= \int_{0}^{\infty} e^{- i {{(\frac{y}{\sqrt{π} (2 n + 1)} - \sqrt{π} (2 n + 1))}^{2} - π {(2 n + 1)}^{2}}} dy$ $= e^{i π {(2 n + 1)}^{2}} \int_{0}^{\infty} e^{- i {\frac{y}{\sqrt{π} (2 n + 1)} - \sqrt{π} (2 n + 1)}^{2}} dy$ $=_{\frac{y}{\sqrt{π} (2 n + 1)} - \sqrt{π} (2 n + 1) = z} e^{i π {(2 n +)}^{2}} \int_{- \sqrt{π} (2 n + 1)}^{+ \infty} e^{- {iz}^{2}} \sqrt{π} (2 n + 1) dz$ $. . . . be continued . . .$
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