How to evaluate this integral

1. Feb 6, 2010

eraserxp

$$\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{\beta(\varepsilon_{F}-\varepsilon_i)}}{\beta}d\beta=?$$

When $$\varepsilon_{F}>\varepsilon_{i}$$, the contour is C1 + C2 (see the attached file). Let $$\beta\rightarrow \infty$$, the integration along C2 vanishes. Then the result is given by the value of $$e^{\beta(\varepsilon_{F}-\varepsilon_i)}/\beta$$ at pole $$\beta=0$$, which is 1.

The problem is that I don't understand why the contribution from C2 vanishes when $$\beta$$ approaches infinity. It seems to me that
$$\int_{C^2}\frac{e^{\beta(\varepsilon_{F}-\varepsilon_i)}}{\beta}d\beta \leq \lim _{\beta\rightarrow \infty}\left|\frac{e^{\beta(\varepsilon_{F}-\varepsilon_i)}}{\beta}\right|\lim_{\beta\rightarrow \infty} \left|d\beta\right|_{\gamma-i\infty}^{\gamma+i\infty}=\infty\cdot\infty$$
The above equation doesn't rule out the possibility for the integration to be zero, but it still confuses me.

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2. Feb 6, 2010

torquil

Seems a bit fishy to me, but I'm not expert. I can see that the contributions to the C2 integral from large negative values of beta will approach zero, but the contributions close to the Im(z) axis far away from the origin seem to not be damped enough (only one power of beta in the denominator).

Your expressions containing limits are a bit confusing, since you say that beta -> infinity. I guess you meant |beta| -> infinity. Also, I think you meant to say the "by the residue of [integrand] at pole beta=0 which is 1".

Also, I'm not sure what the meaning is of the expression where you take the absolute value of the integration measure d\beta.

Torquil

3. Feb 6, 2010

eraserxp

Thanks!

I should state it more clearly. Beta here is a complex number, and |beta|--> infinity. My last limit is confusing. It is the length of arc C2 at |beta|--> infinity, which is $$\lim_{R\rightarrow \infty}\pi R$$.

The integral was taken from a book, and it was said in that book that when $$\varepsilon_{F}>\varepsilon_{i}$$, the result is 1. I have been thinking about it for a long time, but couldn't understand it. I guess maybe what the book says is wrong, and there is a typo in the book. It seems to me that $$\varepsilon_{F}$$ should be smaller than $$\varepsilon_{i}$$.

Let's consider the case when $$\varepsilon_{F}<\varepsilon_{i}$$

According to Cachy integral formula, we have

$$f(a)=\frac{1}{2\pi i}\oint_{C}\frac{f(z)}{z-a}dz$$ .

$$e^{0(\varepsilon_{F}-\varepsilon_i)}=\frac{1}{2\pi i}\int_{C^1+C^2}\frac{e^{\beta(\varepsilon_{F}-\varepsilon_i)}}{\beta}d\beta=\frac{1}{2\pi i}\int_{C^1}\frac{e^{\beta(\varepsilon_{F}-\varepsilon_i)}}{\beta}d\beta +\frac{1}{2\pi i}\int_{C^2}\frac{e^{\beta(\varepsilon_{F}-\varepsilon_i)}}{\beta}d\beta$$.

In the above equation, the first integration on the right is
$$\frac{1}{2\pi i}\left|\int_{C^1}g(\beta)d\beta \right| \leq \lim_{R\rightarrow\infty}\frac{1}{2\pi i}\left|\frac{e^{\beta(\varepsilon_{F}-\varepsilon_i)}}{\beta} \right|_{\mbox{max}}\left| \int_{C^1}d\beta\right|\sim 0*\infty =0$$,
the second integration is the integral that we seek to evaluate, therefore,
$$\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{\beta(\varepsilo n_{F}-\varepsilon_i)}}{\beta}d\beta=g(0)=1$$

I am not sure whether the above derivation is correct or not, please help check it. Thank you!

Last edited: Feb 6, 2010
4. Feb 7, 2010

torquil

You need $$\varepsilon_{F}>\varepsilon_{i}$$ so that the exponential kills the contributions on the far left of the contour.

What confuses me is that the power of beta in the denominator is not large enough to suppress the value of the intregral along C2, as far as I can see. Near the imaginary axis, the exponential doesn't suppress it. So I actually believe that the integral doesn't converge,but I may be wrong.

Btw, I don't think it is well-defined to write 0*infinity = 0 the way you do in your limit calculation.

If your expression is related to physics (epsilon_F sounds to me like the fermi energy?), then you could maybe find this calculation somewhere in the physics literature.

Torquil

5. Feb 7, 2010

Count Iblis

See "[URL [Broken] of Jordan´s lemma[/URL], you can generalize it slightly so that it can be applied to this problem.

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6. Feb 7, 2010

eraserxp

Thank you!

I know how to evaluate the line integral now.

$$\beta -\gamma=R e^{i\theta}$$

$$d\beta =d(\beta-\gamma)=i\cdot R \cdot e^{i\theta}d\theta$$

$$\int_{C^1}\frac{e^{\beta(\varepsilon_{F}-\varepsilon_i)}}{\beta}d\beta=\int_{\pi/2}^{3\pi/2}\frac{e^{(\gamma + R e^{i\theta})(\varepsilon_{F}-\varepsilon_i)}}{\gamma + R e^{i\theta}}iR e^{i\theta}d\theta=\int_{\pi/2}^{3\pi/2}\frac{e^{(\gamma + R \cos\theta + i R \sin\theta)(\varepsilon_{F}-\varepsilon_i)}}{\gamma + R \cos\theta + i R \sin\theta}iR e^{i\theta}d\theta$$

$$\leq \int_{\pi/2}^{3\pi/2}\left|\frac{e^{(\gamma + R \cos\theta + i R \sin\theta)(\varepsilon_{F}-\varepsilon_i)}}{\gamma + R \cos\theta + i R \sin\theta}iR e^{i\theta}\right|d\theta = \int_{\pi/2}^{3\pi/2}\frac{e^{\gamma (\varepsilon_{F}-\varepsilon_i)}e^{R\cos\theta(\varepsilon_{F}-\varepsilon_i)}}{\sqrt{\gamma^2+R^2+2R\gamma\cos\theta}}R d\theta$$

As $$R\rightarrow \infty$$, the denominator in the integrand will cancels with R. Since $$\pi/2 \leq \theta \leq 3\pi/2$$, $$\cos\theta\$$ is negative. If $$\varepsilon_{F}-\varepsilon_i>0$$, then the numerator in the integrand will decay to 0 as R approaches infinity.

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7. Feb 7, 2010

eraserxp

I think I understand the situation here. The complex number $$\beta$$ not only includes the information about the magnitude but also the information about the phase $$\theta$$, so it is not a good way to look at the limit of

$$\frac{e^{\beta(\varepsilon_{F}-\varepsilon_{i})}}{\beta}$$

from the purely magnitude sense, like

$$\lim_{\beta\rightarrow \infty}\left|\frac{e^{\beta(\varepsilon_{F}-\varepsilon_{i}})}{\beta}\right|$$

It is better to split the real and imaginary parts of beta out, and then focus on the limit of the real part, like I did in the last post.

Your speculation is correct. This expression is related to physics, $$\varepsilon_{F}$$ is the Fermi energy.