Convergence of Improper Integrals with Imaginary Units

[matematikuvol]

Homework Statement

Solve integrals

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

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

Homework Equations

\int e^x=e^x+C

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?

 

[Dickfore]

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

EDIT:
Hint:
<br /> e^{x + i y} = e^{x} \left(\cos{y} + i \sin{y}\right)<br />

 

[matematikuvol]

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

 

[Dickfore]

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

 

[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 :(

 

[Dickfore]

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

 

[matematikuvol]

Can you write second step?

 

[Dickfore]

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

 

[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?

 

[Dickfore]

Do you know this equivalence

<br /> \lim_{x \rightarrow x_0} f(x) = 0 \Leftrightarrow \lim_{x \rightarrow x_0} \vert f(x) \vert = 0<br />

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:
<br /> \vert f(x) - 0 \vert = \vert f(x) \vert = \vert \vert f(x) \vert \vert = \vert \vert f(x) \vert - 0 \vert<br />

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

What is this limit:
<br /> \lim_{x \rightarrow -\infty} {e^{a x}} = ?, \; a &gt; 0<br />

 

[matematikuvol]

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

 

[Dickfore]

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