- #1

Hummingbird25

- 86

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## Homework Statement

Looking at the Integral

[tex]a_n = \int_{0}^{\pi} \frac{sin(x)}{x+n\pi}[/tex]

prove that [tex]a_n \geq a_{n+1}[/tex]

## Homework Equations

## The Attempt at a Solution

Here is my now proof:

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

[tex]$\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[/tex]

Common denominator:

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

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

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.