Integral Homework Help: Evaluating \int \frac{1-cos(x)}{x^2} dx | Basic Methods

[Jory]

Homework Statement

Evaluate $$\int$$ $$\frac{1-cos(x)}{x^2}$$ dx
given that it exists (The integral is between 0 and Inf, couldn't figure out how to add limits)

Homework Equations

No idea, looking round, it seems this is related to the Dirichlet Integral but the proof is way beyond the level I'm at. Given that this is from an exam that seems weird. Is there a way to compute this with more rudimentary methods, perhaps by using the complex plane?

The Attempt at a Solution

None

 

[Tangent87]

Do you know how to do the integral:

$$\int_{-\infty}^{\infty}\frac{1-e^{ix}}{x^2}dx ?$$

Can you see the connection?

 

[elabed haidar]

no actually how did you come with that? please explain

 

[Tangent87]

Because $$e^{ix}=cosx+isinx$$ and so your integral is just the real part of my integral. What level of maths are you at? Because if you don't know that identity your unlikely to know the complex methods that will allow you to do this integral.

 

[elabed haidar]

owwwwwwwwwwwwwwwwwww sorry for asking this stupid question

 

[Tangent87]

Also forgot to mention your integrand is EVEN so you can write the integral as:

$$\frac{1}{2}Re\left(\int_{-\infty}^{\infty}\frac{1-e^{ix}}{x^2}dx\right)$$.

Hopefully you know what to do now.

 

[Jory]

Just to point out, that wasn't the OP ;), but yeah that helps. Will take another look

Edit:

So can you calculate the residue at 0 using the Laurent expansion of e^ix, which gives the answer as Pi?

Thanks

 

[Tangent87]

Just to point out, that wasn't the OP ;), but yeah that helps. Will take another look

Edit:

So can you calculate the residue at 0 using the Laurent expansion of e^ix, which gives the answer as Pi?

Thanks

Oh yeah, sorry lol. Yes that's exactly right so your integral (0 to infinity) will be pi/2

 

[Jory]

Awesome

Thanks bud

 

[Jory]

Hmm I'm still not getting the correct answer.

Using the Taylor expansion of e^ix, then dividing by x^2, gives me a coefficient of -i in front of the z^-1 term does it not?

Which makes the Residue = 2Pi, then dividing this by 2 gives me Pi, not Pi/2 as the answer.

Where am I going wrong?

 

[Tangent87]

Because the residue should be pi, not 2pi because when you close the semi-circular contour in the upper half plane you only go round the origin on an infinitesimal semi-circle so it's pi, it would be 2pi if we went all around the origin on a full circle but we don't in this case. (It might help to draw the contour).