- #1
Hummingbird25
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This is a repost, the reason I feared that people who miss the the original.
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]
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]$=\pi\int_0^\pi\left({\sin t\over (t^2+(2n+1)\pi t+(n^2+n)\pi)}\right)\,dt[/tex]
From here we can use the fact that the denominator on the half of the interval where sinus is negative is larger than the denominator of each and every other corresponding point on the other half of the interval, so the whole integral must be positive.
q.e.d.
How does it look now?
Sincerely Maria.
Homework Statement
Looking at the Integral
[tex]a_n = \int_{0}^{\pi} \frac{sin(x)}{x+n\pi}[/tex]
prove that [tex]a_n \geq a_{n+1}[/tex]
Homework Equations
The Attempt at a Solution
Here is my now proof:
the difference between the two integrals, we seek to show:
[tex]$\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.