Homework Help: Miscellaneous Definite Integrals

1. Jun 6, 2012

lostminty

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

show that

$\int^{∞}_{0}\frac{sin^{2}x}{x^{2}}dx= \frac{\pi}{2}$

2. Relevant equations

consider

$\oint_{C}\frac{1-e^{i2z}}{z^{2}}dz$

where C is a semi circle of radius R, about 0,0 with an indent (another semi circle) excluding 0,0.

3. The attempt at a solution

curve splits into
C1, line segment from -R to -ε
C2, semi circle z=εe θ goes from ∏ to 0
C3, line segment from ε to R
C4, semi circle z=Re θ goes from 0 to ∏

function is holomorphic in/on C. so integral =0

for C4, applying limit R > infinity integral = 0

C3. really stuck here

I sub in the value of z

to get

$\int^{0}_{∏}\frac{1-e^{i2e^{iθ}}}{ε^{2}e^{2iθ}}εe^{iθ}$

but I can't take the limit ε -> 0 of this since theres an ε on the denominator

2. Jun 6, 2012

vela

Staff Emeritus
You forgot an ε in the exponent and a factor of i.
$$\int^0_\pi \frac{1-e^{i2\varepsilon e^{i\theta}}}{\varepsilon^2 e^{2i\theta}} i\varepsilon e^{i\theta}\,d\theta$$

3. Jun 6, 2012

lostminty

Thank you. I was half awake when I wrote that.

I just don't know how to get that ε out from the denominator, I cannot take the limit otherwise.

4. Jun 6, 2012

vela

Staff Emeritus
Try expanding the exponential and the numerator as a series.

5. Jun 6, 2012

lostminty

The teacher gave me a hint, use L'hopitals Rule, which isn't clear to be possible since i forgot to add in the limit

6. Jun 6, 2012

vela

Staff Emeritus
I'm not sure what you mean by that. In any case, either way should work. The numerator is first-order in ε. Along with the factor of ε from dz, it cancels the ε2 in the denominator, so you will get a finite result.

7. Jun 7, 2012

lostminty

I mean, differentiating the top and bottom before taking the limit.

as in a taylor series?

8. Jun 7, 2012

vela

Staff Emeritus
I still have no clue what you mean.