Fresnel whole-line moment from the damped complex Gaussian

Write $u_0=\mathrm{Re}\alpha >0$. For finite $R$, the substitution $y=\sqrt{u_0}x$ in [F1] gives ${\int }_{-R}^R{e}^{-u_0{x}^2}dx=u_0^{-1\mathrm{/}2}{\int }_{-\sqrt{u_0}R}^{\sqrt{u_0}R}{e}^{-y^2}dy$; letting $R\to \mathrm{\infty }$ and using [F17] shows that the real Gaussian majorant has finite one-sided integrals. Since $\mathrm{\mid }{e}^{-\alpha x^2}\mathrm{\mid }=e^{-u_0{x}^2}$ by [F20], [F42] gives absolute convergence of each real and imaginary tail; [F37] and [F39] specify these one-sided limits and [F38] defines their two-sided sum $F(\alpha )$. By [F18], every finite-interval complex integral exists; the cut-additivity clause of [F15], applied componentwise at $0$, shows $F_R(\alpha )={\int }_{-R}^R{e}^{-\alpha x^2}dx\to F(\alpha )$. On $S_R=[-R,R{]}^2$, the integrand is continuous by Euler's formula [F20], differentiability and continuity of sine, cosine and the real exponential [F21], [F22], [F25], and polynomial continuity [F32]; also $e^{-\alpha (x^2+y^2)}=e^{-\alpha x^2}{e}^{-\alpha y^2}$ by [F19]. Its real and imaginary parts are integrable by [F14]. Expanding the complex product into four real products, applying [F16] to each, using the complex-coordinate formulas [F35], and using linearity [F15] yields $F_R(\alpha {)}^2={\int }_{S_R}{e}^{-\alpha (x^2+y^2)}d(x,y)$.

Put $D_s=\{(x,y):x^2+y^2\le s^2\}$ and $A_{\epsilon ,R}=D_R\setminus \mathrm{int}{D}_{\epsilon }$. On $K_{+}=[\epsilon ,R]\times [0,\pi ]$ and $K_{-}=[\epsilon ,R]\times [-\pi ,0]$, the polar map $P(r,\theta )=(r\cos \theta ,r\sin \theta )$ has determinant $r$ by [F21] and [F5]. Each rectangle has an open rectangular neighborhood with positive radial coordinate and angular width strictly below $2\pi$. If two points there have the same image, equal Euclidean norms give equal radii; after division by that radius, their trigonometric pairs agree. [F2] then gives sine of the angle difference zero and cosine equal to one. By [F3], the difference is $m\pi$; repeated use of [F4], with evenness from [F5] when $m<0$, gives cosine $(-1{)}^m$, so $m$ is even. The neighborhood's angular width below $2\pi$ forces the difference to be zero. Thus each map is injective on a neighborhood and its derivative is invertible there because the determinant is positive. By [F30], every point in the upper half-annulus has a canonical polar angle in $[0,2\pi )$; if the second coordinate is positive, [F6], [F5], and [F3] rule out an angle in $(\pi ,2\pi )$, while if it is zero [F3] leaves only $0$ or $\pi$. For a lower-half point with negative second coordinate, its canonical angle is in $(\pi ,2\pi )$ by the same sign facts, and subtracting $2\pi$ places it in $(-\pi ,0)$ without changing its trigonometric pair; points with zero second coordinate use angle $0$ or $-\pi$. These arguments prove that the images are exactly the upper and lower closed half-annuli. For $s>0$, their curved boundary arcs are graphs of $\pm \sqrt{s^2-x^2}$; the radicand is continuous by [F32], and for nonnegative $u,v$ one has $\mathrm{\mid }\sqrt{u}-\sqrt{v}{\mathrm{\mid }}^2\le \mathrm{\mid }u-v\mathrm{\mid }$ by [F26] and algebra, proving continuity of those graphs. Each radial segment is a constant-function graph after exchanging coordinates if needed. Each graph has content zero by [F8]; their finite unions do as well by combining finite covers from [F9]. Thus the half-annuli are bounded with content-zero boundary and hence Jordan by [F11]; they are compact by [F10]. Now apply [F1] separately to the real and imaginary parts on each. By componentwise Fubini [F16], each transformed integral is $\pi {\int }_{\epsilon }^R{e}^{-\alpha r^2}rdr$. Writing $\alpha =A+iB$ and differentiating the Euler representation with [F20]--[F24] gives $(e^{-\alpha r^2}{)}^{\mathrm{\prime }}=-2\alpha r{e}^{-\alpha r^2}$; componentwise [F29] therefore evaluates this radial integral as $(e^{-\alpha {\epsilon }^2}-e^{-\alpha R^2})\mathrm{/}(2\alpha )$. Their intersection is two radial segments, content-zero by the same graph argument. On a bounding rectangle, the zero extensions of the two half-annulus integrands sum to the annulus integrand off those segments and to twice it on them; [F12] removes this content-zero difference, while linearity [F15] adds the componentwise integrals. Thus ${\int }_{A_{\epsilon ,R}}{e}^{-\alpha (x^2+y^2)}d(x,y)=\pi (e^{-\alpha {\epsilon }^2}-e^{-\alpha R^2})\mathrm{/}\alpha$. The same circle-graph argument makes $D_{\epsilon }$ compact Jordan; [F14] integrates the continuous real and imaginary parts there. For $h=e^{-\alpha (x^2+y^2)}$, [F20] and [F33] give $\mathrm{\mid }h\mathrm{\mid }=e^{-u_0(x^2+y^2)}\le 1$; the containing square has area $4{\epsilon }^2$ by [F34], and [F13], [F15], [F40] give $\mathrm{\mid }{\int }_{D_{\epsilon }}h\mathrm{\mid }\le {\int }_{D_{\epsilon }}\mathrm{\mid }h\mathrm{\mid }\le 4{\epsilon }^2$. On the containing square, the zero extensions over $D_{\epsilon }$ and $A_{\epsilon ,R}$ sum to $h$ off the boundary circle and to $2h$ on it; [F12] removes this content-zero difference and linearity [F15] gives ${\int }_{D_R}h={\int }_{D_{\epsilon }}h+{\int }_{A_{\epsilon ,R}}h$. Letting $\epsilon \downarrow 0$ therefore gives ${\int }_{D_R}h=\pi (1-e^{-\alpha R^2})\mathrm{/}\alpha$.

Set $E_R=S_R\setminus \mathrm{int}{D}_R$. It is closed and bounded, hence compact by [F10]; its boundary is contained in $\mathrm{\partial }{S}_R\cup \mathrm{\partial }{D}_R$. The square boundary is a finite union of line graphs and the circle boundary is the pair of continuous semicircle graphs proved in step 1.2, so [F8] and [F9] make this union content zero and [F11] makes $E_R$ Jordan. Both $E_R$ and $D_R$ are compact Jordan subsets of $S_R$, so their zero extensions are integrable by [F13] and [F14]. On $S_R$ these extensions sum to $h$ off $\mathrm{\partial }{D}_R$ and to $2h$ on the content-zero circle; [F12] and linearity [F15] therefore give ${\int }_{E_R}h+{\int }_{D_R}h={\int }_{S_R}h$. Every point of $E_R$ has $x^2+y^2\ge R^2$, so $\mathrm{\mid }x\mathrm{\mid }\ge R\mathrm{/}\sqrt{2}$ or $\mathrm{\mid }y\mathrm{\mid }\ge R\mathrm{/}\sqrt{2}$. Put $G={\int }_{\mathbb{R}}{e}^{-u_0{x}^2}dx<\mathrm{\infty }$ and $H(t)={\int }_{\mathrm{\mid }x\mathrm{\mid }\ge t}{e}^{-u_0{x}^2}dx$; $H(t)\to 0$ by [F37] and the convergent nonnegative improper Gaussian integral [F17]. By [F19], the majorant factors as $e^{-u_0{x}^2}{e}^{-u_0{y}^2}$. On finite bounding rectangles, [F13], [F15], and [F16] bound its integral over the vertical and horizontal tail strips by $2H(R\mathrm{/}\sqrt{2})G$. For $h(x,y)=e^{-\alpha (x^2+y^2)}$, [F40] bounds the complex integral's modulus by the integral of $\mathrm{\mid }h\mathrm{\mid }$, so it is at most $2H(R\mathrm{/}\sqrt{2})G$. The integrand is continuous on the compact Jordan set $E_R$, so its components are integrable by [F14]. Passing to the one-dimensional Gaussian tail limits preserves this bound. It tends to zero, so the square and disc integrals have the same limit. Also $\mathrm{\mid }{e}^{-\alpha R^2}\mathrm{\mid }=e^{-u_0{R}^2}\to 0$: [F33] makes this nonnegative function decreasing in $R$, and a positive limiting lower bound would contradict the finite Gaussian integral [F1], [F17]. Steps 1.1 and 1.2, with $e^{-\alpha R^2}\to 0$, now give $F(\alpha {)}^2=\pi \mathrm{/}\alpha$.

The function $F$ is continuous on the right half-plane. Fix a compact neighborhood with $\mathrm{Re}\alpha ,\mathrm{Re}\beta \ge c_0>0$. On $[-R,R]$, differentiating $e^{-(\beta +t(\alpha -\beta ))x^2}$ in $t$ by [F20]--[F24] and applying [F29] gives $\mathrm{\mid }{e}^{-\alpha x^2}-e^{-\beta x^2}\mathrm{\mid }\le R^2\mathrm{\mid }\alpha -\beta \mathrm{\mid }$. By [F40], the finite-interval contribution to $\mathrm{\mid }F(\alpha )-F(\beta )\mathrm{\mid }$ is therefore at most $2R^3\mathrm{\mid }\alpha -\beta \mathrm{\mid }$. Each omitted tail is bounded by ${\int }_{\mathrm{\mid }x\mathrm{\mid }>R}{e}^{-c_0{x}^2}dx$, which tends to zero as $R\to \mathrm{\infty }$ by [F37], [F39], and the scaled Gaussian integral [F1], [F17]; choosing $R$ and then $\mathrm{\mid }\alpha -\beta \mathrm{\mid }$ proves continuity. To justify the branch continuously, write $\alpha =x+iy$, $q=\sqrt{x^2+y^2}$, $u=\sqrt{(q+x)\mathrm{/}2}>0$, and $v=y\mathrm{/}(2u)$. For nonnegative $s,t$, $\mathrm{\mid }\sqrt{s}-\sqrt{t}{\mathrm{\mid }}^2\le \mathrm{\mid }s-t\mathrm{\mid }$, so [F26] and [F32] make $q,u,v$ continuous. Also $2uv=y$ and $u^2-v^2=((q+x{)}^2-y^2)\mathrm{/}(2(q+x))=x$ since $q^2=x^2+y^2$, hence $(u+iv{)}^2=\alpha$. Let $\theta =\arg \alpha \in (-\pi ,\pi ]$ from [F31]. Since $x=\mathrm{\mid }\alpha \mathrm{\mid }\cos \theta >0$, $\cos \theta >0$. On $[-\pi ,-\pi \mathrm{/}2]$, [F4] writes $\cos \theta =\sin (\theta +\pi \mathrm{/}2)\le 0$ by [F5], [F6]; on $[\pi \mathrm{/}2,\pi ]$, put $t=\theta -\pi \mathrm{/}2\in [0,\pi \mathrm{/}2]$ and use period $2\pi$, oddness, and [F6] to get $\cos \theta =\sin (\pi +t)=\sin (t-\pi )=-\sin (\pi -t)\le 0$. Hence $-\pi \mathrm{/}2<\theta <\pi \mathrm{/}2$. Then the specified square root has positive real part because its argument is $\theta \mathrm{/}2\in (-\pi \mathrm{/}4,\pi \mathrm{/}4)$ and cosine is positive there by [F4], [F6]. Both it and $u+iv$ square to $\alpha$; since complex numbers form a field [F35], any two roots differ only by sign, and their positive real parts force equality. Thus $H(\alpha )=F(\alpha )(u+iv)\mathrm{/}\sqrt{\pi }$ is continuous; steps 2.1 and $(u+iv{)}^2=\alpha$ give $H(\alpha {)}^2=(F(\alpha {)}^2\mathrm{/}\pi )\alpha =1$. On any line segment, if the endpoint values of $H$ were $1$ and $-1$, [F7] applied to its continuous real part would force that part to vanish somewhere, impossible for a value in $\{-1,1\}$. Therefore $H$ is constant along every segment. At $\alpha =1$, [F17] gives $H(1)=1$, so $F(\alpha )=\sqrt{\pi }{\alpha }^{-1\mathrm{/}2}$.

Fix $\lambda >0$ and $\sigma =\pm 1$, and for $\epsilon >0$ set $\alpha =\epsilon -i\sigma \lambda \mathrm{/}2$. For any $\epsilon \ge 0$ and $T>R>0$, [F27] gives the phase derivative and [F28] justifies integration by parts on $[R,T]$. The amplitude derivative is $(e^{-\epsilon u^2}\mathrm{/}u{)}^{\mathrm{\prime }}=-2\epsilon e^{-\epsilon u^2}-e^{-\epsilon u^2}\mathrm{/}{u}^2$ by [F22]--[F24]; the boundary terms are at most $2\mathrm{/}(\lambda R)$. For $\epsilon >0$, using $u\ge R$ and then [F22], [F23], [F29], and [F36] gives ${\int }_R^{\mathrm{\infty }}2\epsilon e^{-\epsilon u^2}du\le R^{-1}{\int }_R^{\mathrm{\infty }}2\epsilon u{e}^{-\epsilon u^2}du=e^{-\epsilon R^2}\mathrm{/}R\le 1\mathrm{/}R$; when $\epsilon =0$ this derivative term is zero. Also $e^{-\epsilon u^2}\le 1$ by [F33], so ${\int }_R^{\mathrm{\infty }}{e}^{-\epsilon u^2}{u}^{-2}du\le 1\mathrm{/}R$. Including the factor $1\mathrm{/}\lambda$ gives $\mathrm{\mid }{\int }_R^T{e}^{-\epsilon u^2}{e}^{i\sigma \lambda u^2\mathrm{/}2}du\mathrm{\mid }\le 4\mathrm{/}(\lambda R)$, and the negative tail has the same bound by $u\mapsto -u$. For $\epsilon =0$, [F40] makes each real and imaginary tail satisfy the Cauchy condition in [F41]; hence the one-sided limits exist, and [F38] defines the ordinary two-sided improper Fresnel integral. The estimate is uniform for $\epsilon \ge 0$; on every finite interval the damped integrand converges uniformly as $\epsilon \downarrow 0$. Given a tolerance, first choose $R$ so both tails are uniformly small and then pass to the limit on $[-R,R]$; this proves that the Abel limit equals the ordinary two-sided improper integral. As $\epsilon \downarrow 0$, $\mathrm{\mid }\alpha \mathrm{\mid }\to \lambda \mathrm{/}2$ and $\arg \alpha \to -\sigma \pi \mathrm{/}2$, so step 3.1 gives $\sqrt{\pi }(\lambda \mathrm{/}2{)}^{-1\mathrm{/}2}{e}^{i\sigma \pi \mathrm{/}4}=\sqrt{2\pi \mathrm{/}\lambda }{e}^{i\sigma \pi \mathrm{/}4}$.