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Question Number 9275 by geovane10math last updated on 27/Nov/16

About the Euler - Mascheroni Constant :

γ = \int_{0}^{1} \frac{1}{1 - x} + \frac{1}{\ln x} d x

We can see that

1 - x \neq 0 \Leftrightarrow 1 \neq x;

x = 0 \to \ln x ∄ .

If x \neq 0 and x \neq 1, in the Cartesian Plane,

this function has singularity x = 0 a n d x = 1 .

So, I could write

f (x) = \frac{1}{1 - x} + \frac{1}{\ln x}

\int_{0}^{1} f (x) d x = \lim_{A \to 0^{+}} \lim_{B \to 1^{-}} \int_{A}^{B} f (x) d x ?

P S : S o r r y b y m y w o r s e E n g l i s h

Commented by geovane10math last updated on 27/Nov/16

O r

\int_{0}^{1} f (x) d x = \int_{0}^{0, 5} f (x) d x + \int_{0, 5}^{1} f (x) d x

= \lim_{A \to 0^{+}} \int_{A}^{0, 5} f (x) d x + \lim_{B \to 1^{-}} \int_{0, 5}^{B} f (x) d x

Both ways are right ? ? ? ? ? ? ?

Answered by 123456 last updated on 28/Nov/16

you should divide it in two

suppose c = 1 / 2 (any value 0 < c < 1 can be used)

\underset{0}{\int^{1}} f (x) d x = \lim_{a \to 0^{+}} \underset{a}{\int^{c}} f (x) d x + \lim_{a \to 1^{-}} \underset{c}{\int^{b}} f (x) d x

the two part must converge in order to

it converge .

\lim_{A \to 0^{+}} \lim_{B \to 1^{-}} \underset{A}{\int^{B}} f (x) d x look fine too

Commented by geovane10math last updated on 28/Nov/16

T h a n k s!