"When you differentiate it again, you did not get the function." Wrong, you do get it. The derivative is 1/(x ln(x) ) + sum_(n=1)^infty (ln x)^(n-1) / (x n!). Write the ln(x)^(n-1) as ln(x)^n / ln(x) and take the factor 1/(x ln(x) ) out of the whole sum, then you have 1/(x ln(x) ) times (1 + sum_(n=1)^infty (ln x)^n / n! ) = 1/(x ln(x) ) times sum_(n=0)^infty (ln x)^n / n! = 1/(x ln(x) ) times e^(ln(x)) = 1/(x ln(x) ) timex x = 1/ln(x). It works.
Integral of e^x / x dx is a special "Ei(x)" function
Ei(ln(x)) + C is the answer
Yep, but if you want to express it in an “exact” form, the one in the video is one of it
Therefore, we can prove that li(x) = ln(ln(x)) + ∑(n=1, ∞) (((ln(x))^n)/(n*n!)).
Do you edit using python manim?
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absolutely
Yes
What this problem is solved by Integration by parts or not????
When you differentiate it again, you did not get the function.
Because the integral of 1/lnx doesn't exist in closed form . We have to do it by expansion only
"When you differentiate it again, you did not get the function."
Wrong, you do get it. The derivative is 1/(x ln(x) ) + sum_(n=1)^infty (ln x)^(n-1) / (x n!). Write the ln(x)^(n-1) as ln(x)^n / ln(x) and take the factor 1/(x ln(x) ) out of the whole sum, then you have
1/(x ln(x) ) times (1 + sum_(n=1)^infty (ln x)^n / n! ) = 1/(x ln(x) ) times sum_(n=0)^infty (ln x)^n / n!
= 1/(x ln(x) ) times e^(ln(x)) = 1/(x ln(x) ) timex x = 1/ln(x).
It works.
@@bjornfeuerbacher5514 sorry, u r right
Omg this is little to me even i though in my math
What
What
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Классно 😊😊😊
А почему только 2 лайка😢😢😢