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lf l cannot reach a number in my intergration such as... \(\int_{2}^{infinity} \frac{1}{2n-1}\)

 

Could l replace infinty with a variable and make a limit approaching infinity?

HighSchoolCalculus  Feb 15, 2017
 #1
avatar+92806 
0

Hi HSC

That is because this integral does not converge.

 

\(\int_{2}^{\infty} \frac{1}{2n-1}dn\\ =\left[\frac{ln(2n-1)}{2}\right]_2^\infty\\ =\left[\frac{ln(2n-1)}{2}\right]_2^\infty\\ =\infty-\frac{ln3}{2}\\ =\infty\)

 

back up by wolfram|alpha

https://www.wolframalpha.com/input/?i=integra+of+1%2F(2x-1)dx+from+x%3D2+to+x%3Dinfty

 

I was surprised by this because here is the graph...

 

The area of the green section represents the integral.... it looks like it converges doesn't it....

 

Maybe Heureka or Alan or some other person might like to comment ://

 

Melody  Feb 15, 2017

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