# Computing integrals on the half line

1. Feb 16, 2012

### hunt_mat

Hi,

In my fluids work I have come to integrals of the type:
$$\int_{0}^{\infty}\frac{e^{ikx}}{ak^{2}+bk+c}dk$$
I was thinking of evaluating this via residue calculus but I can't think of the right contour, any suggestions?

Mat

Last edited: Feb 16, 2012
2. Feb 17, 2012

### Some Pig

Try the punctured disc with boundary $C_{\epsilon}\cup[\epsilon,R]\cup C_R.$

#### Attached Files:

• ###### pdisc.jpg
File size:
2.3 KB
Views:
125
3. Feb 17, 2012

### Dickfore

Notice that $\vert e^{i \, k \, x} \vert = e^{-x \, \mathrm{Im}k}$. This means that the integral would diverge when we take the circle at infinity on the lower (upper) semicircle for positive (negative) x.

4. Feb 17, 2012

### Dickfore

Notice that the inverse Fourier transform of the Heaviside step function:
$$\int_{-\infty}^{\infty}{\frac{d k}{2\pi} \, \theta(k) \, e^{i \, k \, x}} = -\frac{1}{2\pi \, i \, x}, \ \mathrm{Im}x > 0$$
Thus, we may represent the Heaviside step function as:
$$\theta(k) = -\frac{1}{2\pi \, i} \, {d t \, \frac{e^{-i \, k \, t}{t + i \, \eta}}, \ \eta \rightarrow +0$$

Why do we need it? Because your integral goes to:
$$\int_{-\infty}^{\infty}{f(k) \, e^{i \, k \, x} \, \theta(k)}$$
If you substitute the integral representation for the step function and change the order of integration, you should get:
$$-\frac{1}{2\pi \, i} \, \int_{-\infty}^{\infty}{\frac{d t}{t + i \, \eta} \, \int_{-\infty}{\infty}{f(k) \, e^{i \, k \, (x - t)}}}$$
Now, you may use the residue theorem for the integral over k, but you need to close the contour in different half-planes, depending on whetgher $x > t$ or $x < t$. The remaining integral over t is again over the whole real line, but , due to the above conditions, should be split into $-\infty$ to x, and from x to $\infty]. Then, making a sub 5. Feb 20, 2012 ### hunt_mat I should point out that [itex]x\in\mathbb{R}$

Dick, can you explain the substitution again, I don't quite get what you're doing here and you still haven't mentioned the contour you're integrating over.