Calculating the Fourier integral representation of f(x)

[J6204]

Homework Statement

Considering the function $$f(x) = e^{-x}, x>0$$ and $$f(-x) = f(x)$$. I am trying to find the Fourier integral representation of f(x).

Homework Equations

$$f(x) = \int_0^\infty \left( A(\alpha)\cos\alpha x +B(\alpha) \sin\alpha x\right) d\alpha$$

$$A(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\cos\alpha u du$$

$$B(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\sin\alpha u du$$

The Attempt at a Solution

$$A(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\cos\alpha u du = \frac{1}{\pi} \int_0^\infty e^{-u}\cos \alpha u du$$
**Notice here how I used 0 and iinfinity as my bounds, is this correct?**
Below is the calculation of this integral, I will save myself sometime and just post the answer I got.
$$A(\alpha) = \frac{1}{\alpha ^2 +1}$$
Calculting $$B(\alpha),$$

$$B(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\sin\alpha u du = \frac{1}{\pi} \int_0^\infty e^{-u}\sin \alpha u du$$

**Notice again here how I used 0 and infinity as my bounds, is this correct?**
Below is the calculation of this integral, I will save myself sometime and just post the answer I got.
$$B(\alpha) = \frac{\alpha}{\alpha^2 +1}$$

 

[Mark44]

Homework Statement

Considering the function $$f(x) = e^{-x}, x>0$$ and $$f(-x) = f(x)$$. I am trying to find the Fourier integral representation of f(x).

Wouldn't this be the same as saying $f(x) = e^{-|x|}$, for $x \in \mathbb R$?

Homework Equations

$$f(x) = \int_0^\infty \left( A(\alpha)\cos\alpha x +B(\alpha) \sin\alpha x\right) d\alpha$$

$$A(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\cos\alpha u du$$

$$B(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\sin\alpha u du$$

The Attempt at a Solution

$$A(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\cos\alpha u du = \frac{1}{\pi} \int_0^\infty e^{-u}\cos \alpha u du$$
**Notice here how I used 0 and iinfinity as my bounds, is this correct?**

Hard to say, since you just posted the value you got after you found after integrating.

Below is the calculation of this integral, I will save myself sometime and just post the answer I got.
$$A(\alpha) = \frac{1}{\alpha ^2 +1}$$
Calculting $$B(\alpha),$$

$$B(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\sin\alpha u du = \frac{1}{\pi} \int_0^\infty e^{-u}\sin \alpha u du$$

**Notice again here how I used 0 and infinity as my bounds, is this correct?**

Same comment as above.

Below is the calculation of this integral, I will save myself sometime and just post the answer I got.
$$B(\alpha) = \frac{\alpha}{\alpha^2 +1}$$

 

[J6204]

Wouldn't this be the same as saying $f(x) = e^{-|x|}$, for $x \in \mathbb R$?
Hard to say, since you just posted the value you got after you found after integrating.
Same comment as above.

I checked my answer on wolfram alpha and these calculations were correct. I actually did forget to include the $$\frac{1}{pi}$$ that was outside of the integral. Am I correct in using 0 to infinity to my bounds or should it be -infinity to infinity?

 

[Mark44]

Am I correct in using 0 to infinity to my bounds or should it be -infinity to infinity?

The latter, I believe, but it's not clear to me how you are defining your function.
In your problem statement, you say
$f(x) = e^{-x}, x > 0$ and then $f(-x) = f(x)$. The second part makes no sense if x > 0.
Also, $e^{-x}$ is defined at x = 0, so there's no reason not to include it in the interval.

 

[J6204]

The latter, I believe, but it's not clear to me how you are defining your function.
In your problem statement, you say
$f(x) = e^{-x}, x > 0$ and then $f(-x) = f(x)$. The second part makes no sense if x > 0.
Also, $e^{-x}$ is defined at x = 0, so there's no reason not to include it in the interval.

okay so below is the calculation of the integral but then how would I find the definite integral using -infinity and infinity?

$$A(\alpha) = \frac{1}{\pi}\int f(u)\cos\alpha u du = \frac{1}{\pi} \int e^{-u}\cos \alpha u du = \frac{1}{\pi } \frac{e^{-x}(\alpha sin(\alpha x)- (cos(\alpha x ))}{\alpha ^2 + 1}$$

when I tried to plug it into a online calculator it could not calculate the definite integgral

 

[Mark44]

Your problem is still not clear to me as to what the domain of your function f is. You didn't clear up the uncertainty I expressed in post #4:

In your problem statement, you say
$f(x) = e^{-x}, x > 0$ and then $f(-x) = f(x)$. The second part makes no sense if x > 0.
Also, $e^{-x}$ is defined at x = 0, so there's no reason not to include it in the interval.

Whether to use limits of 0 and $\infty$ or $-\infty$ and $\infty$ is directly connected to what the domain of your function is.

when I tried to plug it into a online calculator it could not calculate the definite integgral

Just use reasonably large numbers. I don't think there will be much difference between an upper limit of 1000 versus an upper limit of 10,000.

 

[J6204]

Your problem is still not clear to me as to what the domain of your function f is. You didn't clear up the uncertainty I expressed in post #4:

Whether to use limits of 0 and $\infty$ or $-\infty$ and $\infty$ is directly connected to what the domain of your function is.
Just use reasonably large numbers. I don't think there will be much difference between an upper limit of 1000 versus an upper limit of 10,000.

I am a little confused of the domain also. The only states that the function is f(x) = e^{-x} , x> 0 and f(-x) = f(x)

 

[Mark44]

I am a little confused of the domain also. The only states that the function is f(x) = e^{-x} , x> 0 and f(-x) = f(x)

In that case, I think the problem is asking for the Fourier integral representation of $f(x) = e^{-|x|}$.
On the interval $(-\infty, 0], f(x) = e^x$, and on the interval $[0, \infty), f(x) = e^{-x}$. I don't see any reason not to include 0 in each of these intervals, since $e^0 = 1$.
To evaluate $\int_{-\infty}^\infty \dots$ you'll need to break this integral into two parts: one for the left half of the number line, and the other for the right half. Also, you'll need to use limits, as one or the other limit of integration makes each an improper integral.

 

[J6204]

In that case, I think the problem is asking for the Fourier integral representation of $f(x) = e^{-|x|}$.
On the interval $(-\infty, 0], f(x) = e^x$, and on the interval $[0, \infty), f(x) = e^{-x}$. I don't see any reason not to include 0 in each of these intervals, since $e^0 = 1$.
To evaluate $\int_{-\infty}^\infty \dots$ you'll need to break this integral into two parts: one for the left half of the number line, and the other for the right half. Also, you'll need to use limits, as one or the other limit of integration makes each an improper integral.

can I use the fact that it is a even function therefore B will be 0 and I will have to calculate A?

 

[Mark44]

can I use the fact that it is a even function therefore B will be 0 and I will have to calculate A?

Sounds reasonable to me.