- #1
romsofia
- 601
- 323
Is this "legal"?
[tex]{sinx= \sum^\inf_0 \frac{-1^nx^{2n+1}}{(2n+1)!}}[/tex]
Now, let's say we take the integral of this: [tex]{\int sinx = \int \sum^\inf_0 \frac{-1^nx^{2n+1}}{(2n+1)!} = \sum^\inf_0 \frac{-1^n}{(2n+1)!} \int x^{2n+1}}[/tex]
Which we will get: [tex]{\sum^\inf_0 \frac{-1^n}{(2n+1)!} \frac{x^{2n+2}}{2n+2}+C}[/tex]
Which of course is the power series for cosx (as we expected).
The reason why I'm asking this is, am I allowed to make this substitution for sinx/lnx?
I.e: [tex]{\int \frac{sinx}{lnx} = \int \sum^\inf_0 \frac{-1^nx^{2n+1}}{(2n+1)!}*\frac{1}{lnx} = \sum^\inf_0 \frac{-1^n}{(2n+1)!} \int \frac{x^{2n+1}}{lnx} = \sum^\inf_0 \frac{-1^n}{(2n+1)!} Ei((2n+2)lnx)+C}[/tex]
Thanks for your time and help.
EDIT: I think mute might've come to this conclusion a while back (I'd have to check my old threads).
[tex]{sinx= \sum^\inf_0 \frac{-1^nx^{2n+1}}{(2n+1)!}}[/tex]
Now, let's say we take the integral of this: [tex]{\int sinx = \int \sum^\inf_0 \frac{-1^nx^{2n+1}}{(2n+1)!} = \sum^\inf_0 \frac{-1^n}{(2n+1)!} \int x^{2n+1}}[/tex]
Which we will get: [tex]{\sum^\inf_0 \frac{-1^n}{(2n+1)!} \frac{x^{2n+2}}{2n+2}+C}[/tex]
Which of course is the power series for cosx (as we expected).
The reason why I'm asking this is, am I allowed to make this substitution for sinx/lnx?
I.e: [tex]{\int \frac{sinx}{lnx} = \int \sum^\inf_0 \frac{-1^nx^{2n+1}}{(2n+1)!}*\frac{1}{lnx} = \sum^\inf_0 \frac{-1^n}{(2n+1)!} \int \frac{x^{2n+1}}{lnx} = \sum^\inf_0 \frac{-1^n}{(2n+1)!} Ei((2n+2)lnx)+C}[/tex]
Thanks for your time and help.
EDIT: I think mute might've come to this conclusion a while back (I'd have to check my old threads).
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