Using Contour Integration with no singularity

[QFT25]

Homework Statement

Using contour methods, evaluate the following integrals. In any case in which you wish to argue that some portion of a closed contour gives a negligible contribution, you should explain why that is so.

Integral[E^I(k+delta*I)x^2 dx from negative Infinity to Infinity]

as Delta tends toward positive zero.

Homework Equations

Residue theorem which states that the the Integral of a complex function over a closed contour equals 2*Pi*I times the sum of all of it's residues which are determined by singular points.

The Attempt at a Solution

I can't find any singularities in the integrand so I initially thought I can still use a semi circular contour and set the integral over that contour to zero. That way the integral over the real line would equal negative integral of the semi circle. But when I do that I get an exponential to the exponential which I cannot integrate. That leads me to think that my approach is wrong. I never saw an integral like this before so I'm sort of at a lost where to start. Any help will be appreciated. Note this is for a Mathematical Physics class, not a proof based complex analysis class. [/B]

 

[lurflurf]

You do not need a singularity. Integrate on a closed contour with 4 sections. The integral will be zero since the are no singularities.
Integral=Integral1+Integral2+Integral3+Integral4
Choose the contour so that if it is made big two parts approach 0 and the other two establish
$$\int_{-\infty}^\infty \! e^{\imath(k+\delta\imath)x^2}\,\mathrm{d}x=C\int_{-\infty}^\infty \! e^{-\delta x^2}\,\mathrm{d}x$$
For a constant C you will determine in terms of k and delta.
Then try to do the integral without k.
It is well known and can be done using contour integration though other ways are easier.

 

[QFT25]

You do not need a singularity. Integrate on a closed contour with 4 sections. The integral will be zero since the are no singularities.
Integral=Integral1+Integral2+Integral3+Integral4
Choose the contour so that if it is made big two parts approach 0 and the other two establish
$$\int_{-\infty}^\infty \! e^{\imath(k+\delta\imath)x^2}\,\mathrm{d}x=C\int_{-\infty}^\infty \! e^{-\delta x^2}\,\mathrm{d}x$$
For a constant C you will determine in terms of k and delta.
Then try to do the integral without k.
It is well known and can be done using contour integration though other ways are easier.

Would the contour be some type of annulus or look like the contour of a branch cut?

 

[QFT25]

Oh wait what about a rectangle? Is that the contour?

 

[lurflurf]

It is possible to make it a rectangle or other shapes like a trapezoid or two triangles. One obvious one is

$$0=\left(\int_{-\infty}^\infty+\int_{\infty}^{\infty\sqrt{\frac{\delta\imath}{k+\delta\imath}}}+\int_{\infty\sqrt{\frac{\delta\imath}{k+\delta\imath}}}^{-\infty\sqrt{\frac{\delta\imath}{k+\delta\imath}}}+\int_{-\infty\sqrt{\frac{\delta\imath}{k+\delta\imath}}}^{-\infty}\right) \! e^{\imath(k+\delta\imath)x^2}\,\mathrm{d}x$$

 

[QFT25]

It is possible to make it a rectangle or other shapes like a trapezoid or two triangles. One obvious one is

$$0=\left(\int_{-\infty}^\infty+\int_{\infty}^{\infty\sqrt{\frac{\delta\imath}{k+\delta\imath}}}+\int_{\infty\sqrt{\frac{\delta\imath}{k+\delta\imath}}}^{-\infty\sqrt{\frac{\delta\imath}{k+\delta\imath}}}+\int_{-\infty\sqrt{\frac{\delta\imath}{k+\delta\imath}}}^{-\infty}\right) \! e^{\imath(k+\delta\imath)x^2}\,\mathrm{d}x$$

Did you do change of variables to get those bounds? I'm trying to make a rectangle and I am able to show that the vertical legs go to zero and then I get the two horizontal parts equal to each other. f(z)=(z+i*b) where b is a constant. But then I get stuck because I don't know what to pick as b.

 

[lurflurf]

Yes that's it we change variables then argue any infinite limits will do
$$\int_{-\infty}^{\infty} \! e^{-s x^2}\,\mathrm{d}x=$$
$$\frac{1}{\sqrt{s}}\int_{-\infty \sqrt{s}}^{\infty \sqrt{s}} \! e^{-u^2}\,\mathrm{d}u=$$
$$\frac{1}{\sqrt{s}} \left( \int_{-\infty \sqrt{s}}^{-\infty } +\int_{-\infty }^{\infty } +\int_{\infty }^{\infty \sqrt{s}} \right) \! e^{-u^2}\,\mathrm{d}u=$$
$$\frac{1}{\sqrt{s}}\int_{-\infty }^{\infty } \! e^{-u^2}\,\mathrm{d}u$$

so for your problem let s^2=k+delta*I and suitably restricted
argue two of the three integrals with sqrt{s} in the limits are small

 

[Ray Vickson]

Did you do change of variables to get those bounds? I'm trying to make a rectangle and I am able to show that the vertical legs go to zero and then I get the two horizontal parts equal to each other. f(z)=(z+i*b) where b is a constant. But then I get stuck because I don't know what to pick as b.

I will write $a+ib$ instead of $k + i \delta$, so you have $\lim_{R \to \infty} I_R$, where
$$I_R = \int_R^R e^{-(b-ia) x^2} \, dx$$
with $b > 0$. If you change variables to $z = x \sqrt{b-ia}$ then (setting $u-iv = \sqrt{b-ia}$) we have $I_R = 1/\sqrt{b-ia}\: J_R$ where
$$J_R = \int_{-(u-iv) R}^{(u-iv) R} e^{-z^2} \, dz.$$
We can write
$$J_R = A_R+B_R+ \int_{-uR}^{uR} e^{-z^2} \, dz,\\
A_R = \int_{-(u-iv)R}^{-uR} e^{-z^2} \, dz, \\ B_R = \int_{uR}^{(u+iv)R} e^{-z^2} \, dz$$.
In $A_R$ we have $z = -Ru -iy, 0 \leq y \leq R|v|$. Thus
$$|e^{-z^2}|= |e^{-R^2 u^2 + y^2} e^{-2Ruyi}| \leq |e^{-(R^2u^2-R^2 v^2)}| = e^{-R^2 b},$$
so $|A_R| \leq R|v| e^{-b R^2} \to 0$ as $R \to \infty$. Similarly, $|B_R| \to 0$. Thus, the answer is
$$\text{answer} = \frac{1}{\sqrt{b-ia}} \int_{-\infty}^{\infty} e^{-z^2} \, dz.$$

There no need to look for a closed contour; just deforming the 1-dimensional contour is good enough.