# Using Contour Integration with no singularity

1. Nov 4, 2016

### QFT25

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.

3. 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.

2. Nov 5, 2016

### 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.

3. Nov 5, 2016

### QFT25

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

4. Nov 5, 2016

### QFT25

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

5. Nov 5, 2016

### lurflurf

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

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

6. Nov 5, 2016

### QFT25

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.

7. Nov 7, 2016

### lurflurf

Yes thats 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

8. Nov 7, 2016

### Ray Vickson

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.

Last edited: Nov 7, 2016