# Integral test by comparison(Please verify my proof)

1. Feb 26, 2008

### Hummingbird25

Integral test by comparison(Please look at my work)

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

Looking at the Integral

$$a_n = \int_{0}^{\pi} \frac{sin(x)}{x+n\pi}$$

prove that $$a_n \geq a_{n+1}$$

2. Relevant equations

3. The attempt at a solution

Proof

given the integral test of comparison and since a_n is convergent, then a_n will always be larger than a_(n+1), by comparisons test.

q.e.d.

Is this surficient? Or do I need to add something that they converge to different limit point?

Sincerely Yours
Maria

Last edited: Feb 27, 2008
2. Feb 26, 2008

### PingPong

What can you say about the integrand of $a_{n+1}$ as compared to that of $a_{n}$ on the interval $(0,\pi)$?

3. Feb 26, 2008

### Hummingbird25

Here is my now proof:

the difference between the two integrals, we seek to show:

$$\forall n\in\mathbb{N}:\int_0^\pi\left({\sin t\over t+n\pi}-{\sin t\over t+(n+1)\pi}\right)\,dt\ge 0$$

Common denominator:

$$=\int_0^\pi\left({\sin t((t+(n+1)\pi)-(t+n\pi))\over (t+n\pi)(t+(n+1)\pi)}\right)\,dt$$

$$=\pi\int_0^\pi\left({\sin t\over (t^2+(2n+1)\pi t+(n^2+n)\pi)}\right)\,dt$$

From here we can use the fact that the denominator on the half of the interval where sinus is negative is larger than the denominator of each and every other corresponding point on the other half of the interval, so the whole integral must be positive.

q.e.d.

How does it look now?

Sincerely Maria.

Last edited: Feb 27, 2008