Menu Close

About-the-Euler-Mascheroni-Constant-0-1-1-1-x-1-ln-x-dx-We-can-see-that-1-x-0-1-x-x-0-ln-x-If-x-0-and-x-1-in-the-Cartesian-Plane-this-fun

Tinku Tara 0 Comments
Pin




Question Number 9275 by geovane10math last updated on 27/Nov/16
About the Euler-Mascheroni Constant: γ = ∫_0 ^1 (1/(1 − x)) + (1/(ln x)) dx We can see that 1− x ≠ 0 ⇔ 1 ≠ x ; x = 0 → ln x ∄ . If x ≠ 0 and x ≠ 1, in the Cartesian Plane, this function has singularity x=0 and x=1. So, I could write f(x) = (1/(1 − x)) + (1/(ln x)) ∫_0 ^1 f(x) dx = lim_(A→0^+ ) lim_(B→1^− ) ∫_A ^B f(x) dx ? PS: Sorry by my worse English
$$\mathrm{About}\:\mathrm{the}\:\mathrm{Euler}-\mathrm{Mascheroni}\:\mathrm{Constant}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\gamma\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}\:−\:{x}}\:+\:\frac{\mathrm{1}}{\mathrm{ln}\:{x}}\:{dx} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{see}\:\mathrm{that}\: \\ $$$$\:\mathrm{1}−\:{x}\:\neq\:\mathrm{0}\:\Leftrightarrow\:\mathrm{1}\:\neq\:{x}\:; \\ $$$${x}\:=\:\mathrm{0}\:\rightarrow\:\mathrm{ln}\:{x}\:\nexists\:. \\ $$$$\mathrm{If}\:{x}\:\neq\:\mathrm{0}\:\mathrm{and}\:{x}\:\neq\:\mathrm{1},\:\mathrm{in}\:\mathrm{the}\:\mathrm{Cartesian}\:\mathrm{Plane}, \\ $$$$\mathrm{this}\:\mathrm{function}\:\mathrm{has}\:\mathrm{singularity}\:{x}=\mathrm{0}\:{and}\:{x}=\mathrm{1}. \\ $$$$\mathrm{So},\:\mathrm{I}\:\mathrm{could}\:\mathrm{write} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}}\:+\:\frac{\mathrm{1}}{\mathrm{ln}\:{x}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right)\:{dx}\:=\:\underset{{A}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\underset{{B}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\int_{{A}} ^{{B}} {f}\left({x}\right)\:{dx}\:\:? \\ $$$${PS}:\:{Sorry}\:{by}\:{my}\:{worse}\:{English} \\ $$
Commented by geovane10math last updated on 27/Nov/16
Or ∫_0 ^1 f(x) dx = ∫_0 ^(0,5) f(x) dx + ∫_(0,5) ^1 f(x) dx = lim_(A→0^+ ) ∫_A ^(0,5) f(x) dx + lim_(B→1^− ) ∫_(0,5) ^B f(x) dx Both ways are right???????
$${Or}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right)\:{dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{0},\mathrm{5}} {f}\left({x}\right)\:{dx}\:+\:\int_{\mathrm{0},\mathrm{5}} ^{\mathrm{1}} {f}\left({x}\right)\:{dx} \\ $$$$=\:\underset{{A}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\int_{{A}} ^{\mathrm{0},\mathrm{5}} {f}\left({x}\right)\:{dx}\:+\:\underset{{B}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\int_{\mathrm{0},\mathrm{5}} ^{{B}} {f}\left({x}\right)\:{dx} \\ $$$$ \\ $$$$\mathrm{Both}\:\mathrm{ways}\:\mathrm{are}\:\mathrm{right}??????? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
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) ∫_0 ^1 f(x)dx=lim_(a→0^+ ) ∫_a ^c f(x)dx+lim_(a→1^− ) ∫_c ^b f(x)dx the two part must converge in order to it converge. lim_(A→0^+ ) lim_(B→1^− ) ∫_A ^B f(x)dx look fine too
$$\mathrm{you}\:\mathrm{should}\:\mathrm{divide}\:\mathrm{it}\:\mathrm{in}\:\mathrm{two} \\ $$$$\mathrm{suppose}\:{c}=\mathrm{1}/\mathrm{2}\:\left(\mathrm{any}\:\mathrm{value}\:\mathrm{0}<{c}<\mathrm{1}\:\mathrm{can}\:\mathrm{be}\:\mathrm{used}\right) \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right){dx}=\underset{{a}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\underset{{a}} {\overset{{c}} {\int}}{f}\left({x}\right){dx}+\underset{{a}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\underset{{c}} {\overset{{b}} {\int}}{f}\left({x}\right){dx} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{part}\:\mathrm{must}\:\mathrm{converge}\:\mathrm{in}\:\mathrm{order}\:\mathrm{to} \\ $$$$\mathrm{it}\:\mathrm{converge}. \\ $$$$\underset{\mathrm{A}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\underset{{B}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\underset{{A}} {\overset{{B}} {\int}}{f}\left({x}\right){dx}\:\mathrm{look}\:\mathrm{fine}\:\mathrm{too} \\ $$
Commented by geovane10math last updated on 28/Nov/16
Thanks!
$${Thanks}! \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *