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Question Number 223580 by Nicholas666 last updated on 30/Jul/25
∫_0 ^(1 ) ((e^(−r^2 ) sin(1/r^2 )ln(r+1))/r^2 ) dr
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}\:} \:\frac{{e}^{−\boldsymbol{{r}}^{\mathrm{2}} } \boldsymbol{\mathrm{sin}}\left(\mathrm{1}/\boldsymbol{{r}}^{\mathrm{2}} \right)\boldsymbol{\mathrm{ln}}\left(\boldsymbol{{r}}+\mathrm{1}\right)}{\boldsymbol{{r}}^{\mathrm{2}} }\:\boldsymbol{\mathrm{d}{r}} \\ $$$$ \\ $$
Answered by MathematicalUser2357 last updated on 31/Jul/25
does not exist.
$$\mathrm{does}\:\mathrm{not}\:\mathrm{exist}. \\ $$
Commented by yerlow last updated on 31/Jul/25
I think its possible however, im still answering.
Answered by yerlow last updated on 01/Aug/25
This integral does not converge, But i know how to integrate this, even if it diverges. Let the integral be defined as: lim_(ε→0^+ ) ∫_ε ^1 ((e^(−r^2 ) ln(r+1))/r^2 )sin((1/r^2 ))dr If the limit cancels, then the oscillation of the integral “cancels” In the distributional sense. u=(1/r^2 )⇒r=(1/( (√u))), dr = −(1/2)u^(−(3/2)) du Substituting the variables above to the integral.. ∫_∞ ^1 ⌊e^(−(1/u)) ln(1+(1/( (√u))))sin(u)∙(1/2)u^(−(3/2)) ∙(1/u)du =(1/2)∫_1 ^∞ e^(−(1/u)) ln(1+(1/( (√u))))u^(−(5/2)) sin(u) du This integral does converge! Now since the integral cannot be expressed in any function at all, I will be using Series Manipulation. e^(−(1/u)) =Σ_(n=0) ^∞ (((−1)^n )/(n!))∙(1/u^n ) ln(1+(1/( (√u))))=Σ_(m=1) ^∞ (((−1)^(m+1) )/m)∙(1/u^(m/2) ) Define These series expansions as n and m then substitute into the integral ∫_1 ^∞ ⌊n⌋∙⌊m⌋∙u^(−(5/2)) ∙sin(u) du Expand n amd m as a double sum and combine all powers of u: Σ_(n=0) ^∞ Σ_(m=1) ^∞ (((−1)^(n+m+1) )/(n!∙m))∫_1 ^∞ ((sin(u))/u^(n+(m/2)+(5/2)) )du Define the intgral on the right above as I_(n,m) So.. I=Σ_(n=0) ^∞ Σ_(m=1) ^∞ (((−1)^(n+m+1) )/(n!+m)) I_(n,m) ✓
$$\mathrm{This}\:\mathrm{integral}\:\mathrm{does}\:\mathrm{not}\:\mathrm{converge}, \\ $$$$\mathrm{But}\:\mathrm{i}\:\mathrm{know}\:\mathrm{how}\:\mathrm{to}\:\mathrm{integrate} \\ $$$$\mathrm{this},\:\mathrm{even}\:\mathrm{if}\:\mathrm{it}\:\mathrm{diverges}. \\ $$$$ \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{be}\:\mathrm{defined}\:\mathrm{as}: \\ $$$$\underset{\varepsilon\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\int_{\varepsilon} ^{\mathrm{1}} \frac{{e}^{−{r}^{\mathrm{2}} } \mathrm{ln}\left({r}+\mathrm{1}\right)}{{r}^{\mathrm{2}} }\mathrm{sin}\left(\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\right){dr} \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{cancels},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{oscillation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{integral}\:“\mathrm{cancels}'' \\ $$$$\mathrm{In}\:\mathrm{the}\:\mathrm{distributional}\:\mathrm{sense}. \\ $$$${u}=\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\Rightarrow{r}=\frac{\mathrm{1}}{\:\sqrt{{u}}},\:{dr}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}{u}^{−\frac{\mathrm{3}}{\mathrm{2}}} {du} \\ $$$$\mathrm{Substituting}\:\mathrm{the}\:\mathrm{variables}\:\mathrm{above}\:\mathrm{to}\:\mathrm{the}\:\mathrm{integral}.. \\ $$$$\int_{\infty} ^{\mathrm{1}} \lfloor{e}^{−\frac{\mathrm{1}}{{u}}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{u}}}\right)\mathrm{sin}\left({u}\right)\centerdot\frac{\mathrm{1}}{\mathrm{2}}{u}^{−\frac{\mathrm{3}}{\mathrm{2}}} \centerdot\frac{\mathrm{1}}{{u}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} {e}^{−\frac{\mathrm{1}}{{u}}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{u}}}\right){u}^{−\frac{\mathrm{5}}{\mathrm{2}}} \mathrm{sin}\left({u}\right)\:{du} \\ $$$$\mathrm{This}\:\mathrm{integral}\:\mathrm{does}\:\mathrm{converge}! \\ $$$$\mathrm{Now}\:\mathrm{since}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{in}\:\mathrm{any}\:\mathrm{function}\:\mathrm{at}\:\mathrm{all}, \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{be}\:\mathrm{using}\:\mathrm{Series}\:\mathrm{Manipulation}. \\ $$$$ \\ $$$${e}^{−\frac{\mathrm{1}}{\mathrm{u}}} =\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}!}\centerdot\frac{\mathrm{1}}{{u}^{\mathrm{n}} } \\ $$$$\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{u}}}\right)=\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}+\mathrm{1}} }{{m}}\centerdot\frac{\mathrm{1}}{{u}^{\frac{{m}}{\mathrm{2}}} } \\ $$$$\mathrm{Define}\:\mathrm{These}\:\mathrm{series}\:\mathrm{expansions}\:\mathrm{as}\:{n}\:\mathrm{and}\:{m}\:\mathrm{then}\:\mathrm{substitute}\:\mathrm{into}\:\mathrm{the}\:\mathrm{integral} \\ $$$$\int_{\mathrm{1}} ^{\infty} \lfloor{n}\rfloor\centerdot\lfloor{m}\rfloor\centerdot{u}^{−\frac{\mathrm{5}}{\mathrm{2}}} \centerdot\mathrm{sin}\left({u}\right)\:{du} \\ $$$$\mathrm{Expand}\:{n}\:\mathrm{amd}\:{m}\:\mathrm{as}\:\mathrm{a}\:\mathrm{double} \\ $$$$\mathrm{sum}\:\mathrm{and}\:\mathrm{combine}\:\mathrm{all}\:\mathrm{powers}\:\mathrm{of} \\ $$$${u}: \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+{m}+\mathrm{1}} }{{n}!\centerdot{m}}\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{sin}\left({u}\right)}{{u}^{{n}+\frac{{m}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{2}}} }{du} \\ $$$$\mathrm{Define}\:\mathrm{the}\:\mathrm{intgral}\:\mathrm{on}\:\mathrm{the}\:\mathrm{right}\:\mathrm{above}\:\mathrm{as}\:{I}_{{n},{m}} \\ $$$$\mathrm{So}.. \\ $$$$\mathrm{I}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+{m}+\mathrm{1}} }{{n}!+{m}}\:{I}_{{n},{m}} \checkmark \\ $$
Answered by Tawa11 last updated on 31/Jul/25

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