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During my research I ran into the following type of an oscillatory integral, for some values of nonzero reals $a,b$:

$f(R):=\int_{0}^{R} e^{2 \pi i (ar^2 + br)} dr$

and I am interested in finding a good estimate for it when $R>0$ is large, in terms of $R$ and the two parameters $a,b$. I expect $\frac{1}{R}f(R)$ to converge as $R \to \infty$ uniformly in $b$, but a more quantitative estimate will be helpful. I guess this is classical but since I know very little in this area I'm not sure where to find such results.

EDIT: Thanks for your help. In the end I just used a version of van der Corput's lemma, but the answers offered here were still interesting to read.

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Any knowledge about $a$ and $b$? – Fabian Jul 9 '12 at 20:38
If $a\ne0$ is real or $\Im a>0$ the integral from zero to $\infty$ converges. – Andrew Jul 9 '12 at 20:41
But the asymptotics might depend on the sign of the parameters... So it would be helpful to know more about the signs to give the appropriate asymptotic expression. – Fabian Jul 9 '12 at 20:42
@Fabian: any two real numbers, both nonzero. – Mark Jul 9 '12 at 20:55

Use the Stationary Phase Method.

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I'm not sure if this answer is correct. Please feel free to modify it...

In a first step, we perform a variable substitution $r=Rx$ in order to eliminate $R$ from the integration bounds. We have $$f(R) = R \int_0^1\!dx\,e^{2\pi i (a R^2 x^2 + b R x)} = R \,\tilde f(R).$$

As the integrand is holomorphic, we are allowed to deform the integration contour in the complex plain $z = x + i y$. The stationary point is at $z= -b/2aR$.

As the integral starts at $z=0$, we choose a contour on which the phase does not change. This path is (approximately) given by $x=\pm y$ where $\pm$ corresponds to the sign of $ab$.

Discussing the case with $ab> 0$:

Case 1: (a>0, b>0) From the contribution close to $z=0$, we thus have with $z= e^{i \pi/4} t$ (for $a>0$) $$\tilde f(R) \sim e^{i\pi/4} \int_0^c\;dt\,e^{-2\pi a R^2 t^2} \sim\frac{e^{i\pi/4}}{2\sqrt{2a}}. $$

Similar, for $z=1$, we choose $z=1+e^{i\pi/4} t$ and we have a contribution from the end of the contour with the value $$\tilde f(R) \sim -e^{i\pi/4} e^{2\pi i a R^2} \int_c^0\;dt\,e^{-2\pi a R^2 t^2} \sim-\frac{e^{i\pi/4}e^{2\pi i a R^2}}{2\sqrt{2a}}.$$

In total, we have $$f(R) \sim R \frac{e^{i\pi/4}}{2\sqrt{2a}} (1-e^{2\pi i a R^2}). $$

Case 2: (a<0, b<0)

Everything goes along the line of Case 1, we just need to integrate in the direction with $z=e^{-i\pi/4}t$ with the end-result $$f(R) \sim R \frac{e^{-i\pi/4}}{2\sqrt{2|a|}} (1-e^{2\pi i a R^2}). $$

Case 3: $a b \leq 0$

In this case, we have to take the saddle point at $z=-b/2aR$ into account (as the integration contour passes through it. The integral is dominated from the stationary point and the asymptotic result reads using $z=-b/2aR + e^{\mathop{\rm sgn}(a)i\pi/4} t$ $$ f(R) \sim R e^{i b^2\pi/2a} e^{\mathop{\rm sgn}(a)i\pi/4} \int_{-c}^c e^{-2\pi|a| R^2 t^2} \sim e^{i b^2\pi/2a} \frac{e^{\mathop{\rm sgn}(a)i\pi/4} }{2 \sqrt{|a|}}.$$

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You can change variables from $r$ to $r + {b \over a}$ and the integral is $$e^{-i{b^2 \over 4a}}\int_{-{b \over a}}^{R - {b \over a}} e^{2\pi ia r^2}\,dr$$ I'll do the $a > 0$ case, as the $a < 0$ case is similar. Change variables to $\sqrt{a}r$ now and you get $${1 \over \sqrt{a}}e^{-i{b^2 \over 4a}}\int_{-{b \over \sqrt{a}}}^{R\sqrt{a} - {b \over \sqrt{a}}} e^{2\pi i r^2}\,dr$$ $$= {1 \over \sqrt{a}}e^{-i{b^2 \over 4a}}\int_{-{b \over \sqrt{a}}}^{\infty} e^{2\pi i r^2}\,dr - {1 \over \sqrt{a}}e^{-i{b^2 \over 4a}}\int_{R\sqrt{a} - {b \over \sqrt{a}}}^{\infty}e^{2\pi i r^2}\,dr$$ For the second term, one can do repeated integrations by parts to obtain an asymptotic expansion in negative powers of $R\sqrt{a} - {b \over \sqrt{a}}$; one writes $e^{2\pi i r^2} = 4\pi i r e^{2\pi i r^2}\times {1 \over 4\pi i r}$ and integrates the $4\pi i r e^{2\pi i r^2}$. So the integral of the first term will dominate, and the second term will give you explicit asymptotics for the nondominant terms. I'm not sure if there's a way of writing the integral in terms of other better known functions though (but note Robert Israel's comment below.)

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According to Maple, $$\int _{-B}^{\infty }\!{{\rm e}^{2\,i\pi \,{r}^{2}}}{dr}= -\left( 1+i \right) \left( i{\it erfi} \left( \left( 1+i \right) B\sqrt { \pi } \right) -1 \right)/4 $$ – Robert Israel Jul 9 '12 at 22:24

All you need is integration by parts:

$$f(R)=\int_{0}^{R} e^{2 \pi i (ar^2 + br)}dr=\int_{0}^{R}\frac{d e^{2 \pi i (ar^2 + br)}}{2\pi i(2ar+b)}$$

$$=\frac{e^{2 \pi i (aR^2 + bR)}}{2\pi i(2aR+b)}-\frac{1}{2\pi ib}+ \int_{0}^{R}a\frac{ e^{2 \pi i (ar^2 + br)}}{\pi i(2ar+b)^2}dr $$ Let's integrate the remaining integral again by parts:

$$\int_{0}^{R}a\frac{ e^{2 \pi i (ar^2 + br)}}{\pi i(2ar+b)^2}dr= -\int_{0}^{R}a\frac{d e^{2 \pi i (ar^2 + br)}}{2\pi^2 (2ar+b)^3}$$

$$=-\frac{e^{2 \pi i (aR^2 + bR)}}{2\pi^2 (2aR+b)^3}+\frac{1}{2\pi^2 b^3}- \int_{0}^{R}3a^2\frac{ e^{2 \pi i (ar^2 + br)}}{\pi^2 (2ar+b)^4}dr$$ Finally, combine the results together:

$$ f(R)=\frac{e^{2 \pi i (aR^2 + bR)}}{2\pi i(2aR+b)}-\frac{1}{2\pi ib} -\frac{e^{2 \pi i (aR^2 + bR)}}{2\pi^2 (2aR+b)^3}+\frac{1}{2\pi^2 b^3}- \int_{0}^{R}3a^2\frac{ e^{2 \pi i (ar^2 + br)}}{\pi^2 (2ar+b)^4}dr $$ Now, let's estimate the last integral:

$$\left | \int_{0}^{R}3a^2\frac{ e^{2 \pi i (ar^2 + br)}}{\pi^2 (2ar+b)^4}dr \right |\leqslant \frac{3a^2}{\pi^2}\int_{0}^{R}\frac{dr}{\pi^2 (2ar+b)^4}=\frac{a}{2\pi^2b^3}- \frac{a}{2\pi^2(2aR+b)^3}$$ At this point you can decide whether you continue integration by parts or not.

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If $2ar + b = 0$ on the domain of integration, such integrations by parts do not give a valid answer. – Zarrax Jul 10 '12 at 15:26
@Zarrax Good point! – Martin Gales Jul 11 '12 at 4:30

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