Very hard take home test. We're trying to figure out how to integrate:

$\displaystyle f(x) = \int_{1}^{x}\frac{lnt}{1+t^2}dt$

Any help would be much appreciated. Thanks!

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- Nov 28th 2009, 11:09 PMDaveDammitTaylor Polynomial Problem help
Very hard take home test. We're trying to figure out how to integrate:

$\displaystyle f(x) = \int_{1}^{x}\frac{lnt}{1+t^2}dt$

Any help would be much appreciated. Thanks! - Nov 29th 2009, 12:44 AMmr fantastic
- Nov 29th 2009, 01:05 AMsimplependulum
sub $\displaystyle y = \frac{1}{t}$

$\displaystyle dt = - \frac{1}{y^2}~dy$

the integral becomes

$\displaystyle - \int_{\frac{1}{x}}^1 \frac{ - \ln(y) }{1 + y^2 } ~dy$

Since $\displaystyle 0 < \frac{1}{x} \leq y < 1 $

$\displaystyle \frac{1}{1 + y^2} $ can be changed to infinite series form .

$\displaystyle - \sum_{n=0}^{\infty} (-1)^n \int_{\frac{1}{x}}^1 \ln(y) y^{2n}~dy $

Buy by integration by parts

$\displaystyle \int \ln(y) y^{2n}~dy = \ln(y) \frac{y^{2n+1}}{2n+1} - \frac{ y^{2n+1}}{(2n+1)^2 } $

Finally , the integral becomes

$\displaystyle \sum_{n=0}^{\infty} \frac{ (-1)^n}{(2n+1)^2} - \ln(x) \tan^{-1}(\frac{1}{x}) - \sum_{n=0}^{\infty} \frac{ (-1)^n }{ x^{2n+1} (2n+1)^2} $

note that $\displaystyle \tan^{-1}(\frac{1}{x}) $ can be changed to

$\displaystyle \frac{\pi}{2} - \tan^{-1}(x) $

If $\displaystyle x \to \infty $

the integral $\displaystyle \int_1^{\infty} \frac{\ln(x)}{1 + x^2 }~dx $

$\displaystyle = \sum_{n=0}^{\infty} \frac{ (-1)^n}{(2n+1)^2} $ - Nov 29th 2009, 01:55 AMNonCommAlg
**simplependulum**, what you found is not a Taylor series. i think**DaveDammit**is probably looking for the Taylor series of $\displaystyle f(x)$ at $\displaystyle x = 1$. hopefully some day he'll give us more details! - Nov 29th 2009, 09:18 AMDaveDammit
Sorry guys, was working late on the test last night and was very tired when I posted. Pretty cool problems, here's the complete question:

Find the third Taylor polynomial of $\displaystyle f(x) = \int_{1}^{x}\frac{lnt}{1+t^2}dt$ about $\displaystyle a=1$.

Thanks for any help! - Nov 29th 2009, 09:22 AMJameson
- Nov 29th 2009, 09:26 AMhjortur
You do know the definition that a third order Taylor polynomial of f(x) about a is:

$\displaystyle T_3(f(x),a)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3$.

So all you need to do is find the first 3 derivatives of f(x), and sub a=1.

Hope that helps. - Nov 29th 2009, 09:31 AMDaveDammit
I do know the definition of a third order Taylor polynomial. That's not the problem I'm having.

The problem I'm having is using one of the Fundemental Theorems of Calculus to integrate the function with respect to $\displaystyle t$ in order to find my $\displaystyle f(x)$ function. This specific function is fairly challenging for me to integrate. It is a beautiful problem and I will attempt to solve it.

Thanks for your input. - Nov 29th 2009, 09:36 AMJameson
I didn't see NCA's note that simplependulum didn't write a Taylor polynomial so my previous post isn't correct.

I think the point of this problem is you don't need to actually integrate the function because you can use FTCII to find f(1). What is f(1) if the integral of f(t) has bounds from 1 to 1?

From there, f'(a) is just f(t) and all further derivatives can easily be found. You don't need to find the integral of f(t) to write the Taylor polynomial of f(x). - Nov 29th 2009, 09:46 AMhjortur
$\displaystyle

f(x) = \int_{1}^{x}\frac{lnt}{1+t^2}dt

$

Then according to the fundamental theorem of calculus:

$\displaystyle f'(x)=\frac{ln(x)}{1+x^2}$

Do you see why?

You don't need to integrate at all.

Can you finish this one now? - Dec 6th 2009, 03:53 PMDaveDammit
Hey guys,

Sorry I'm so late in responding to this but I just wanted to let you guys know that I completed the problem and I think I did very well on the test. I had to calculate a wicked third derivative of the function but I'm pretty sure I got the answer right.

Thanks for all your help!

Dave