# 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.