# Improper integrals with i

1. Oct 15, 2011

### matematikuvol

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

$$\int^0_{-\infty}e^{(a-ik)x}dx$$

$$\int^{\infty}_{0}e^{-(a+ik)x}dx$$

2. Relevant equations

$$\int e^x=e^x+C$$

3. The attempt at a solution

My troble is with imaginary unit $$i$$

$$\int^0_{-\infty}e^{(a-ik)x}dx=\frac{e^{(a-ik)x}}{a-ik}|^0_{-\infty}$$

$$\int^{\infty}_{0}e^{-(a+ik)x}dx=\frac{e^{-(a+ik)x}}{-a-ik}|^{\infty}_0$$

I don't know how to get results because of $$i$$?

2. Oct 15, 2011

### Dickfore

both integrals converge well if $a > 0$. Just evaluate the limits and do some complex algebra.

EDIT:
Hint:
$$e^{x + i y} = e^{x} \left(\cos{y} + i \sin{y}\right)$$

3. Oct 15, 2011

### matematikuvol

Ok. But here I have x in both terms. Can you show me idea with more details?

4. Oct 15, 2011

### Dickfore

Look at the modulus of the complex number:
$$\left\vert e^{(a + i b) x}\right\vert = e^{a x}$$
When $a < 0$, what does this function tend to as $x \rightarrow \infty$? How about when $x \rightarrow -\infty$? How about when $a > 0$?

5. Oct 15, 2011

### matematikuvol

You tell me if I understand

$$\int^0_{-\infty}e^{(a-ik)x}dx=\frac{e^{(a-ik)x}}{a-ik}|^0_{-\infty}=\frac{e^{ax}}{a-ik}|^0_{-\infty}=\frac{1}{a-ik}$$

$$\int^{\infty}_{0}e^{-(a+ik)x}dx=\frac{e^{-(a+ik)x}}{-a-ik}|^{\infty}_0=\frac{e^{-ax}}{-a-ik}|^{\infty}_0=\frac{1}{a+ik}$$

I still don't understand why I may do that :(

6. Oct 15, 2011

### Dickfore

The end results are correct, but the second step is not.

7. Oct 15, 2011

### matematikuvol

Can you write second step?

8. Oct 15, 2011

### Dickfore

Sure, for the first integral:
$$\frac{e^{0} - \lim_{x \rightarrow -\infty}{e^{(a - i k) x}}}{a - i k}$$

9. Oct 15, 2011

### matematikuvol

Ok. I don't know how to solve that limit. i is not bigger and is not less then zero. How to solve that?

10. Oct 15, 2011

### Dickfore

Do you know this equivalence

$$\lim_{x \rightarrow x_0} f(x) = 0 \Leftrightarrow \lim_{x \rightarrow x_0} \vert f(x) \vert = 0$$

meaning that the limit of some function as $x \rightarrow x_0$ is zero if and only if the limit of the modulus of that function is also zero. It follows from the obvious identity:
$$\vert f(x) - 0 \vert = \vert f(x) \vert = \vert \vert f(x) \vert \vert = \vert \vert f(x) \vert - 0 \vert$$

You should use this rule, as well as the fact that:
$$\vert e^{(a - i k) x} \vert = e^{a x}$$
to show that that limit is zero!

What is this limit:
$$\lim_{x \rightarrow -\infty} {e^{a x}} = ?, \; a > 0$$

11. Oct 16, 2011

### matematikuvol

I don't know that equivalence. How can I show that this is equivalent?

12. Oct 16, 2011

### Dickfore

Doesn't matter. Just show that:
$$\lim_{x \rightarrow -\infty}{\left e^{(a - i k) x} \vert} = 0$$
and use it to show that the limit in the numerator of the fraction is zero.