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Medo
Medo
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Let $0<\theta_1,\theta_2<\pi/2$. Suppose $\psi$ is a smooth real-valued function with compact support.

Consider the oscillatory integral $$I(t):=\int_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx dy.$$ I am trying to find the asymptotic behavior of $I(t)$ as $t\rightarrow \infty$.

Since the phase $x\mapsto t x^2 r^2$ has a non-degenerate stationary point at $x=0$, one is tempted to try the stationary phase method that gives

$$\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx= \frac{C \psi(0)}{\sqrt{t}y}+O(\frac{1}{t y^2}),$$ with some constant $C$. But then we are faced with the critical singularities singularities $y^{-1}$ and $y^{-2}$.

If we use Fubini's theorem then apply the stationary phase method we end up with

$$\int_{0}^{1}\frac{e^{\dot{\imath} t x^2y^2}}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}dr= \frac{\widetilde{C} }{\sqrt{t}x}+O(\frac{1}{t x^2}),$$ which is again not integrable against $\psi$.

Another approach is to change variables $s=\sqrt{t}r$ to write

$$I(t)=\frac{1}{\sqrt{t}} J(t)$$ where $$J(t):=\int_{0}^{\sqrt{t}}\frac{1}{(\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta_1}) (\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy.$$ The idea now is to check whether $\lim_{t\rightarrow \infty}J(t)$ is a constant independent of $t$. But, by Fubini's theorem, we have $$\lim_{t\rightarrow \infty}J(t)=c\int_{\mathbb{R}} \frac{\psi(x)}{x}dx$$ which may diverge.

Let $0<\theta_1,\theta_2<\pi/2$. Suppose $\psi$ is a smooth real-valued function with compact support.

Consider the oscillatory integral $$I(t):=\int_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx dy.$$ I am trying to find the asymptotic behavior of $I(t)$ as $t\rightarrow \infty$.

Since the phase $x\mapsto t x^2 r^2$ has a non-degenerate stationary point at $x=0$, one is tempted to try the stationary phase method that gives

$$\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx= \frac{C \psi(0)}{\sqrt{t}y}+O(\frac{1}{t y^2}),$$ with some constant $C$. But then we are faced with the critical singularities $y^{-1}$ and $y^{-2}$.

If we use Fubini's theorem then apply the stationary phase method we end up with

$$\int_{0}^{1}\frac{e^{\dot{\imath} t x^2y^2}}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}dr= \frac{\widetilde{C} }{\sqrt{t}x}+O(\frac{1}{t x^2}),$$ which is again not integrable against $\psi$.

Another approach is to change variables $s=\sqrt{t}r$ to write

$$I(t)=\frac{1}{\sqrt{t}} J(t)$$ where $$J(t):=\int_{0}^{\sqrt{t}}\frac{1}{(\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta_1}) (\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy.$$ The idea now is to check whether $\lim_{t\rightarrow \infty}J(t)$ is a constant independent of $t$. But, by Fubini's theorem, we have $$\lim_{t\rightarrow \infty}J(t)=c\int_{\mathbb{R}} \frac{\psi(x)}{x}dx$$ which may diverge.

Let $0<\theta_1,\theta_2<\pi/2$. Suppose $\psi$ is a smooth real-valued function with compact support.

Consider the oscillatory integral $$I(t):=\int_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx dy.$$ I am trying to find the asymptotic behavior of $I(t)$ as $t\rightarrow \infty$.

Since the phase $x\mapsto t x^2 r^2$ has a non-degenerate stationary point at $x=0$, one is tempted to try the stationary phase method that gives

$$\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx= \frac{C \psi(0)}{\sqrt{t}y}+O(\frac{1}{t y^2}),$$ with some constant $C$. But then we are faced with the singularities $y^{-1}$ and $y^{-2}$.

If we use Fubini's theorem then apply the stationary phase method we end up with

$$\int_{0}^{1}\frac{e^{\dot{\imath} t x^2y^2}}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}dr= \frac{\widetilde{C} }{\sqrt{t}x}+O(\frac{1}{t x^2}),$$ which is again not integrable against $\psi$.

Another approach is to change variables $s=\sqrt{t}r$ to write

$$I(t)=\frac{1}{\sqrt{t}} J(t)$$ where $$J(t):=\int_{0}^{\sqrt{t}}\frac{1}{(\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta_1}) (\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy.$$ The idea now is to check whether $\lim_{t\rightarrow \infty}J(t)$ is a constant independent of $t$. But, by Fubini's theorem, we have $$\lim_{t\rightarrow \infty}J(t)=c\int_{\mathbb{R}} \frac{\psi(x)}{x}dx$$ which may diverge.

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Iosif Pinelis
Iosif Pinelis
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Let $0<\theta_1,\theta_2<\pi/2$. Suppose $\psi$ is a smooth real-valued function with compact support.

Consider the oscillatory integral $$I(\lambda):=\int_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} \lambda x^2y^2}dx dy.$$$$I(t):=\int_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx dy.$$ I am trying to find the asymptotic behavior of $I(\lambda)$$I(t)$ as $\lambda\rightarrow \infty$$t\rightarrow \infty$.

Since the phase $x\mapsto \lambda x^2 r^2$$x\mapsto t x^2 r^2$ has a non-degenerate stationary point at $x=0$, one is tempted to try the stationary phase method that gives

$$\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} \lambda x^2y^2}dx= \frac{C \psi(0)}{\sqrt{\lambda}y}+O(\frac{1}{\lambda y^2}),$$$$\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx= \frac{C \psi(0)}{\sqrt{t}y}+O(\frac{1}{t y^2}),$$ with some constant $C$. But then we are faced with the critical singularities $y^{-1}$ and $y^{-2}$.

If we use Fubini's theorem then apply the stationary phase method we end up with

$$\int_{0}^{1}\frac{e^{\dot{\imath} \lambda x^2y^2}}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}dr= \frac{\widetilde{C} }{\sqrt{\lambda}x}+O(\frac{1}{\lambda x^2}),$$$$\int_{0}^{1}\frac{e^{\dot{\imath} t x^2y^2}}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}dr= \frac{\widetilde{C} }{\sqrt{t}x}+O(\frac{1}{t x^2}),$$ which is again not integrable against $\psi$.

Another approach is to change variables $s=\sqrt{\lambda}r$$s=\sqrt{t}r$ to write

$$I(\lambda)=\frac{1}{\sqrt{\lambda}} J(\lambda)$$$$I(t)=\frac{1}{\sqrt{t}} J(t)$$ where $$J(\lambda):=\int_{0}^{\sqrt{\lambda}}\frac{1}{(\frac{y}{\sqrt{\lambda}}-e^{\dot{\imath}\theta_1}) (\frac{y}{\sqrt{\lambda}}-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy.$$$$J(t):=\int_{0}^{\sqrt{t}}\frac{1}{(\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta_1}) (\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy.$$ The idea now is to check whether $\lim_{\lambda\rightarrow \infty}J(\lambda)$$\lim_{t\rightarrow \infty}J(t)$ is a constant independent of $\lambda$$t$. But, by Fubini's theorem, we have $$\lim_{\lambda\rightarrow \infty}J(\lambda)=c\int_{\mathbb{R}} \frac{\psi(x)}{x}dx$$$$\lim_{t\rightarrow \infty}J(t)=c\int_{\mathbb{R}} \frac{\psi(x)}{x}dx$$ which may diverge.

Let $0<\theta_1,\theta_2<\pi/2$. Suppose $\psi$ is a smooth real-valued function with compact support.

Consider the oscillatory integral $$I(\lambda):=\int_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} \lambda x^2y^2}dx dy.$$ I am trying to find the asymptotic behavior of $I(\lambda)$ as $\lambda\rightarrow \infty$.

Since the phase $x\mapsto \lambda x^2 r^2$ has a non-degenerate stationary point at $x=0$, one is tempted to try the stationary phase method that gives

$$\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} \lambda x^2y^2}dx= \frac{C \psi(0)}{\sqrt{\lambda}y}+O(\frac{1}{\lambda y^2}),$$ with some constant $C$. But then we are faced with the critical singularities $y^{-1}$ and $y^{-2}$.

If we use Fubini's theorem then apply the stationary phase method we end up with

$$\int_{0}^{1}\frac{e^{\dot{\imath} \lambda x^2y^2}}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}dr= \frac{\widetilde{C} }{\sqrt{\lambda}x}+O(\frac{1}{\lambda x^2}),$$ which is again not integrable against $\psi$.

Another approach is to change variables $s=\sqrt{\lambda}r$ to write

$$I(\lambda)=\frac{1}{\sqrt{\lambda}} J(\lambda)$$ where $$J(\lambda):=\int_{0}^{\sqrt{\lambda}}\frac{1}{(\frac{y}{\sqrt{\lambda}}-e^{\dot{\imath}\theta_1}) (\frac{y}{\sqrt{\lambda}}-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy.$$ The idea now is to check whether $\lim_{\lambda\rightarrow \infty}J(\lambda)$ is a constant independent of $\lambda$. But, by Fubini's theorem, we have $$\lim_{\lambda\rightarrow \infty}J(\lambda)=c\int_{\mathbb{R}} \frac{\psi(x)}{x}dx$$ which may diverge.

Let $0<\theta_1,\theta_2<\pi/2$. Suppose $\psi$ is a smooth real-valued function with compact support.

Consider the oscillatory integral $$I(t):=\int_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx dy.$$ I am trying to find the asymptotic behavior of $I(t)$ as $t\rightarrow \infty$.

Since the phase $x\mapsto t x^2 r^2$ has a non-degenerate stationary point at $x=0$, one is tempted to try the stationary phase method that gives

$$\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx= \frac{C \psi(0)}{\sqrt{t}y}+O(\frac{1}{t y^2}),$$ with some constant $C$. But then we are faced with the critical singularities $y^{-1}$ and $y^{-2}$.

If we use Fubini's theorem then apply the stationary phase method we end up with

$$\int_{0}^{1}\frac{e^{\dot{\imath} t x^2y^2}}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}dr= \frac{\widetilde{C} }{\sqrt{t}x}+O(\frac{1}{t x^2}),$$ which is again not integrable against $\psi$.

Another approach is to change variables $s=\sqrt{t}r$ to write

$$I(t)=\frac{1}{\sqrt{t}} J(t)$$ where $$J(t):=\int_{0}^{\sqrt{t}}\frac{1}{(\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta_1}) (\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy.$$ The idea now is to check whether $\lim_{t\rightarrow \infty}J(t)$ is a constant independent of $t$. But, by Fubini's theorem, we have $$\lim_{t\rightarrow \infty}J(t)=c\int_{\mathbb{R}} \frac{\psi(x)}{x}dx$$ which may diverge.

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Medo
Medo
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Let $0<\theta_1,\theta_2<\pi/2$. Suppose $\psi$ is a smooth real-valued function with compact support.

Consider the oscillatory integral $$I(\lambda):=\int_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} \lambda x^2y^2}dx dy.$$ I am trying to find the asymptotic behavior of $I(\lambda)$ as $\lambda\rightarrow \infty$.

Since the phase $x\mapsto \lambda x^2 r^2$ has a non-degenerate stationary point at $x=0$, one is tempted to try the stationary phase method that gives

$$\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} \lambda x^2y^2}dx= \frac{C \psi(0)}{\sqrt{\lambda}y}+O(\frac{1}{\lambda y^2}),$$ with some constant $C$. But then we are faced with the critical singularities $y^{-1}$ and $y^{-2}$.

If we use Fubini's theorem then apply the stationary phase method we end up with

$$\int_{0}^{1}\frac{e^{\dot{\imath} \lambda x^2y^2}}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}dr= \frac{\widetilde{C} }{\sqrt{\lambda}x}+O(\frac{1}{\lambda x^2}),$$ which is again not integrable against $\psi$.

Another approach is to change variables $s=\sqrt{\lambda}r$ to write

$$I(\lambda)=\frac{1}{\sqrt{\lambda}} J(\lambda)$$ where $$J(\lambda):=\int_{0}^{\sqrt{\lambda}}\frac{1}{(\frac{y}{\sqrt{\lambda}}-e^{\dot{\imath}\theta_1}) (\frac{y}{\sqrt{\lambda}}-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy$$.$$J(\lambda):=\int_{0}^{\sqrt{\lambda}}\frac{1}{(\frac{y}{\sqrt{\lambda}}-e^{\dot{\imath}\theta_1}) (\frac{y}{\sqrt{\lambda}}-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy.$$ The idea now is to check whether $\lim_{\lambda\rightarrow \infty}J(\lambda)$ is a constant independent of $\lambda$. But, by Fubini's theorem, we have $$\lim_{\lambda\rightarrow}J(\lambda)=c\int_{\mathbb{R}} \frac{\psi(x)}{x}dx$$$$\lim_{\lambda\rightarrow \infty}J(\lambda)=c\int_{\mathbb{R}} \frac{\psi(x)}{x}dx$$ which may diverge.

Let $0<\theta_1,\theta_2<\pi/2$. Suppose $\psi$ is a smooth real-valued function with compact support.

Consider the oscillatory integral $$I(\lambda):=\int_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} \lambda x^2y^2}dx dy.$$ I am trying to find the asymptotic behavior of $I(\lambda)$ as $\lambda\rightarrow \infty$.

Since the phase $x\mapsto \lambda x^2 r^2$ has a non-degenerate stationary point at $x=0$, one is tempted to try the stationary phase method that gives

$$\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} \lambda x^2y^2}dx= \frac{C \psi(0)}{\sqrt{\lambda}y}+O(\frac{1}{\lambda y^2}),$$ with some constant $C$. But then we are faced with the critical singularities $y^{-1}$ and $y^{-2}$.

If we use Fubini's theorem then apply the stationary phase method we end up with

$$\int_{0}^{1}\frac{e^{\dot{\imath} \lambda x^2y^2}}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}dr= \frac{\widetilde{C} }{\sqrt{\lambda}x}+O(\frac{1}{\lambda x^2}),$$ which is again not integrable against $\psi$.

Another approach is to change variables $s=\sqrt{\lambda}r$ to write

$$I(\lambda)=\frac{1}{\sqrt{\lambda}} J(\lambda)$$ where $$J(\lambda):=\int_{0}^{\sqrt{\lambda}}\frac{1}{(\frac{y}{\sqrt{\lambda}}-e^{\dot{\imath}\theta_1}) (\frac{y}{\sqrt{\lambda}}-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy$$. The idea now is to check whether $\lim_{\lambda\rightarrow \infty}J(\lambda)$ is a constant independent of $\lambda$. But, by Fubini's theorem, we have $$\lim_{\lambda\rightarrow}J(\lambda)=c\int_{\mathbb{R}} \frac{\psi(x)}{x}dx$$ which may diverge.

Let $0<\theta_1,\theta_2<\pi/2$. Suppose $\psi$ is a smooth real-valued function with compact support.

Consider the oscillatory integral $$I(\lambda):=\int_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} \lambda x^2y^2}dx dy.$$ I am trying to find the asymptotic behavior of $I(\lambda)$ as $\lambda\rightarrow \infty$.

Since the phase $x\mapsto \lambda x^2 r^2$ has a non-degenerate stationary point at $x=0$, one is tempted to try the stationary phase method that gives

$$\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} \lambda x^2y^2}dx= \frac{C \psi(0)}{\sqrt{\lambda}y}+O(\frac{1}{\lambda y^2}),$$ with some constant $C$. But then we are faced with the critical singularities $y^{-1}$ and $y^{-2}$.

If we use Fubini's theorem then apply the stationary phase method we end up with

$$\int_{0}^{1}\frac{e^{\dot{\imath} \lambda x^2y^2}}{(y-e^{\dot{\imath}\theta_1}) (y-e^{\dot{\imath}\theta_2})}dr= \frac{\widetilde{C} }{\sqrt{\lambda}x}+O(\frac{1}{\lambda x^2}),$$ which is again not integrable against $\psi$.

Another approach is to change variables $s=\sqrt{\lambda}r$ to write

$$I(\lambda)=\frac{1}{\sqrt{\lambda}} J(\lambda)$$ where $$J(\lambda):=\int_{0}^{\sqrt{\lambda}}\frac{1}{(\frac{y}{\sqrt{\lambda}}-e^{\dot{\imath}\theta_1}) (\frac{y}{\sqrt{\lambda}}-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy.$$ The idea now is to check whether $\lim_{\lambda\rightarrow \infty}J(\lambda)$ is a constant independent of $\lambda$. But, by Fubini's theorem, we have $$\lim_{\lambda\rightarrow \infty}J(\lambda)=c\int_{\mathbb{R}} \frac{\psi(x)}{x}dx$$ which may diverge.

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Medo
Medo
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Medo
Medo
  • 852
  • 5
  • 13