I have two related questions. The first is: Is there a closed form expression for: int_0^1(1-t)/((t-2)\ln t)dt~~0.507834 I know that there are some very superb integrators on this site. I doubt that a closed form exists, but I may be wrong. If I'm wrong, I'm curious as to how you got your answer! Second question: Is there an elementary function f(t) such that: exp(int_0^1(1-t)/((t-2)ln t)dt)=int_0^1f(t)dt That value is approximately 1.66169. I am very interested in this value. The expression on the left is a great way to express this number. But, I feel like it would be "cleaner" if I could express it simply as an integral, like on the right.

makeupwn

makeupwn

Open question

2022-08-18

0 1 1 t ( t 2 ) ln t d t integral
I have two related questions. The first is: Is there a closed form expression for:
0 1 1 t ( t 2 ) ln t d t 0.507834
I know that there are some very superb integrators on this site. I doubt that a closed form exists, but I may be wrong. If I'm wrong, I'm curious as to how you got your answer!
Second question: Is there an elementary function f ( t ) such that:
exp ( 0 1 1 t ( t 2 ) ln t d t ) = 0 1 f ( t ) d t
That value is approximately 1.66169. I am very interested in this value. The expression on the left is a great way to express this number. But, I feel like it would be "cleaner" if I could express it simply as an integral, like on the right.

Answer & Explanation

Winston Cooper

Winston Cooper

Beginner2022-08-19Added 12 answers

The value of your integral could be expressed as
I = n = 1 ln n 2 n = d d s Φ ( 1 2 , 0 , 0 ) = 1 2 d d s Φ ( 1 2 , 0 , 1 ) ,
where
Φ ( z , s , a ) = n = 1 z n ( n + a ) s
is the Lerch transcendent. The second value you're interested in, surprisingly, has its own name: Somos' quadratic recurrence constant:
exp I = σ = 1 2 3 = 1 1 / 2 2 1 / 4 3 1 / 8
Here's the proof of the series representation. Using that
1 2 t = 1 2 n = 0 ( t 2 ) n = n = 0 t n 2 n + 1 ,
we can write
I = n = 0 1 2 n + 1 0 1 t n + 1 t n ln t d t = n = 0 1 2 n + 1 ln n + 2 n + 1 ,
since
0 1 t n + 1 t n ln t d t = 0 1 ( n n + 1 t y d y ) d t = n n + 1 ( 0 1 t y d t ) d y =
= n n + 1 d y y + 1 = ln n + 2 n + 1 .
Therefore,
I = n = 0 1 2 n + 1 ln n + 2 n + 1 = n = 1 1 2 n ln n + 1 n = 2 n = 1 ln ( n + 1 ) 2 n + 1 n = 1 ln n 2 n = n = 1 ln n 2 n .

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