Mixing
$I(a,b)=int_0^1 sinleft(ln frac{1}{x}right)frac{x^b-x^a}{ln x}dx, b>0,a>0$
$begingroup$
$$I(a,b)=int_0^1 sinleft(ln frac{1}{x}right)frac{x^b-x^a}{ln x}dx, b>0,a>0$$
Please help. I can't find a function to see if the function is absolute and uniform convergent using Weierstrass convergence criteria.
real-analysis
edited Jan 11 at 12:04
Zacky
6,3301858
asked Jan 11 at 11:56
David DanDavid Dan
31
$endgroup$
add a comment |
$begingroup$
$$I(a,b)=int_0^1 sinleft(ln frac{1}{x}right)frac{x^b-x^a}{ln x}dx, b>0,a>0$$
Please help. I can't find a function to see if the function is absolute and uniform convergent using Weierstrass convergence criteria.
real-analysis
edited Jan 11 at 12:04
Zacky
6,3301858
asked Jan 11 at 11:56
David DanDavid Dan
31
$endgroup$
$begingroup$
the question is to calculate the integrals with parameters.I have found this problem in the proposed problems chapter at improper integrals.[drive.google.com/open?id=152R-HdfOJK7Y85YqQUZeVCGHmUOCvz2B] [drive.google.com/open?id=152R-HdfOJK7Y85YqQUZeVCGHmUOCvz2B] are the theorems i usually use for this kind of problem
$endgroup$
– David Dan
Jan 11 at 12:42
$begingroup$
new links in case the ones above doesn't work [imgur.com/ZL15gbO] imgur.com/gallery/MHVcQeE
$endgroup$
– David Dan
Jan 11 at 12:50
$begingroup$
Note that $$sin[log(1/x)]=sin(-log x)=-sinlog x$$
$endgroup$
– clathratus
Jan 11 at 16:57
$begingroup$
Then set $$J(s;a,b)=int_0^1sin(slog x)frac{x^a-x^b}{log x}dx$$ and note that $J(0;a,b)=0$ as well as $J(1;a,b)=I(a,b)$. Then take $frac{partial}{partial s}$ on both sides to see that $$frac{partial}{partial s}J(s;a,b)=int_0^1(x^a-x^b)cos(s log x)dx$$ Which may be easier
$endgroup$
– clathratus
Jan 11 at 17:03
add a comment |
$begingroup$
$$I(a,b)=int_0^1 sinleft(ln frac{1}{x}right)frac{x^b-x^a}{ln x}dx, b>0,a>0$$
Please help. I can't find a function to see if the function is absolute and uniform convergent using Weierstrass convergence criteria.
real-analysis
edited Jan 11 at 12:04
Zacky
6,3301858
asked Jan 11 at 11:56
David DanDavid Dan
31
$endgroup$
$$I(a,b)=int_0^1 sinleft(ln frac{1}{x}right)frac{x^b-x^a}{ln x}dx, b>0,a>0$$
Please help. I can't find a function to see if the function is absolute and uniform convergent using Weierstrass convergence criteria.
real-analysis
real-analysis
edited Jan 11 at 12:04
Zacky
6,3301858
asked Jan 11 at 11:56
David DanDavid Dan
31
edited Jan 11 at 12:04
Zacky
6,3301858
asked Jan 11 at 11:56
David DanDavid Dan
31
edited Jan 11 at 12:04
Zacky
6,3301858
edited Jan 11 at 12:04
Zacky
6,3301858
edited Jan 11 at 12:04
Zacky
6,3301858
6,3301858
asked Jan 11 at 11:56
David DanDavid Dan
31
asked Jan 11 at 11:56
David DanDavid Dan
31
asked Jan 11 at 11:56
David DanDavid Dan
31
31
$begingroup$
the question is to calculate the integrals with parameters.I have found this problem in the proposed problems chapter at improper integrals.[drive.google.com/open?id=152R-HdfOJK7Y85YqQUZeVCGHmUOCvz2B] [drive.google.com/open?id=152R-HdfOJK7Y85YqQUZeVCGHmUOCvz2B] are the theorems i usually use for this kind of problem
$endgroup$
– David Dan
Jan 11 at 12:42
$begingroup$
new links in case the ones above doesn't work [imgur.com/ZL15gbO] imgur.com/gallery/MHVcQeE
$endgroup$
– David Dan
Jan 11 at 12:50
$begingroup$
Note that $$sin[log(1/x)]=sin(-log x)=-sinlog x$$
$endgroup$
– clathratus
Jan 11 at 16:57
$begingroup$
Then set $$J(s;a,b)=int_0^1sin(slog x)frac{x^a-x^b}{log x}dx$$ and note that $J(0;a,b)=0$ as well as $J(1;a,b)=I(a,b)$. Then take $frac{partial}{partial s}$ on both sides to see that $$frac{partial}{partial s}J(s;a,b)=int_0^1(x^a-x^b)cos(s log x)dx$$ Which may be easier
$endgroup$
– clathratus
Jan 11 at 17:03
add a comment |
$begingroup$
the question is to calculate the integrals with parameters.I have found this problem in the proposed problems chapter at improper integrals.[drive.google.com/open?id=152R-HdfOJK7Y85YqQUZeVCGHmUOCvz2B] [drive.google.com/open?id=152R-HdfOJK7Y85YqQUZeVCGHmUOCvz2B] are the theorems i usually use for this kind of problem
$endgroup$
– David Dan
Jan 11 at 12:42
$begingroup$
new links in case the ones above doesn't work [imgur.com/ZL15gbO] imgur.com/gallery/MHVcQeE
$endgroup$
– David Dan
Jan 11 at 12:50
$begingroup$
Note that $$sin[log(1/x)]=sin(-log x)=-sinlog x$$
$endgroup$
– clathratus
Jan 11 at 16:57
$begingroup$
Then set $$J(s;a,b)=int_0^1sin(slog x)frac{x^a-x^b}{log x}dx$$ and note that $J(0;a,b)=0$ as well as $J(1;a,b)=I(a,b)$. Then take $frac{partial}{partial s}$ on both sides to see that $$frac{partial}{partial s}J(s;a,b)=int_0^1(x^a-x^b)cos(s log x)dx$$ Which may be easier
$endgroup$
– clathratus
Jan 11 at 17:03
$begingroup$
the question is to calculate the integrals with parameters.I have found this problem in the proposed problems chapter at improper integrals.[drive.google.com/open?id=152R-HdfOJK7Y85YqQUZeVCGHmUOCvz2B] [drive.google.com/open?id=152R-HdfOJK7Y85YqQUZeVCGHmUOCvz2B] are the theorems i usually use for this kind of problem
$endgroup$
– David Dan
Jan 11 at 12:42
$begingroup$
the question is to calculate the integrals with parameters.I have found this problem in the proposed problems chapter at improper integrals.[drive.google.com/open?id=152R-HdfOJK7Y85YqQUZeVCGHmUOCvz2B] [drive.google.com/open?id=152R-HdfOJK7Y85YqQUZeVCGHmUOCvz2B] are the theorems i usually use for this kind of problem
$endgroup$
– David Dan
Jan 11 at 12:42
$begingroup$
new links in case the ones above doesn't work [imgur.com/ZL15gbO] imgur.com/gallery/MHVcQeE
$endgroup$
– David Dan
Jan 11 at 12:50
$begingroup$
new links in case the ones above doesn't work [imgur.com/ZL15gbO] imgur.com/gallery/MHVcQeE
$endgroup$
– David Dan
Jan 11 at 12:50
$begingroup$
Note that $$sin[log(1/x)]=sin(-log x)=-sinlog x$$
$endgroup$
– clathratus
Jan 11 at 16:57
$begingroup$
Note that $$sin[log(1/x)]=sin(-log x)=-sinlog x$$
$endgroup$
– clathratus
Jan 11 at 16:57
$begingroup$
Then set $$J(s;a,b)=int_0^1sin(slog x)frac{x^a-x^b}{log x}dx$$ and note that $J(0;a,b)=0$ as well as $J(1;a,b)=I(a,b)$. Then take $frac{partial}{partial s}$ on both sides to see that $$frac{partial}{partial s}J(s;a,b)=int_0^1(x^a-x^b)cos(s log x)dx$$ Which may be easier
$endgroup$
– clathratus
Jan 11 at 17:03
$begingroup$
Then set $$J(s;a,b)=int_0^1sin(slog x)frac{x^a-x^b}{log x}dx$$ and note that $J(0;a,b)=0$ as well as $J(1;a,b)=I(a,b)$. Then take $frac{partial}{partial s}$ on both sides to see that $$frac{partial}{partial s}J(s;a,b)=int_0^1(x^a-x^b)cos(s log x)dx$$ Which may be easier
$endgroup$
– clathratus
Jan 11 at 17:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using your definition, we see that there is a natural log in the denominator. An easy way to remove that is simply by using differentiation under the integral. Therefore, we differentiate with respect to $b$ because$$frac {mathrm dx^b}{mathrm db}=x^blog x$$Hence$$begin{align*}mathfrak I'(a,b) & =intlimits_0^1mathrm dx, x^bsinleft(logfrac 1xright)end{align*}$$Now make a transformation $xmapstologtfrac 1x$ to get rid of the nested sine - log function. Therefore, the limits change to zero and infinity, so that$$begin{align*}mathfrak{I}'(a,b) & =intlimits_0^{infty}mathrm dx,e^{x(1+b)}sin x\ & =frac 1{1+(1+b)^2}end{align*}$$Where the last line is derived by integrating the integral by parts twice. Since we have $mathfrak I'(a,b)$, we integrate it back with respect to $b$ to get$$mathfrak{I}(a,b)=arctan(1+b)+C$$To find $C$, we observe that when $a=b$, then $mathfrak{I}(a,b)=0$. Hence, we get that $C=arctan(1+a)$ so the final answer becomes$$intlimits_0^1mathrm dx,frac {x^b-x^a}{log x}sinleft(logfrac 1xright)color{blue}{=arctan(1+b)-arctan(1+a)}$$
answered Jan 11 at 16:16
Frank W.Frank W.
3,5731321
$endgroup$
$begingroup$
How do you make the transformation? Shoudn't the integer after transformation $x->logfrac1x$ become $int_0^infty{sinx}*e^{x(1-b)}dx$. I used the substitution $logfrac1x=u$ , $e^{-u}=x$ then $-frac1xdx=du$. Don't know where I'm wrong. Can you tell me please how to solve $int_0^infty{sinx}*e^{x(1+b)}dx$? And thanks for the help.
$endgroup$
– David Dan
Jan 11 at 18:30
add a comment |
$begingroup$
First, under $xto e^{-x}$
$$ frac{partial I(a,b)}{partial b}=int_0^1sin(ln(frac1x))x^bdx=int_0^infty e^{(b+1)x}sin xdx=frac{1}{(b+1)^2+1} $$
and hence
$$ I(a,b)=int_a^bfrac{1}{(s+1)^2+1}ds=arctan (b+1)-arctan(a+1). $$
answered Jan 11 at 16:07
xpaulxpaul
22.7k24455
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using your definition, we see that there is a natural log in the denominator. An easy way to remove that is simply by using differentiation under the integral. Therefore, we differentiate with respect to $b$ because$$frac {mathrm dx^b}{mathrm db}=x^blog x$$Hence$$begin{align*}mathfrak I'(a,b) & =intlimits_0^1mathrm dx, x^bsinleft(logfrac 1xright)end{align*}$$Now make a transformation $xmapstologtfrac 1x$ to get rid of the nested sine - log function. Therefore, the limits change to zero and infinity, so that$$begin{align*}mathfrak{I}'(a,b) & =intlimits_0^{infty}mathrm dx,e^{x(1+b)}sin x\ & =frac 1{1+(1+b)^2}end{align*}$$Where the last line is derived by integrating the integral by parts twice. Since we have $mathfrak I'(a,b)$, we integrate it back with respect to $b$ to get$$mathfrak{I}(a,b)=arctan(1+b)+C$$To find $C$, we observe that when $a=b$, then $mathfrak{I}(a,b)=0$. Hence, we get that $C=arctan(1+a)$ so the final answer becomes$$intlimits_0^1mathrm dx,frac {x^b-x^a}{log x}sinleft(logfrac 1xright)color{blue}{=arctan(1+b)-arctan(1+a)}$$
answered Jan 11 at 16:16
Frank W.Frank W.
3,5731321
$endgroup$
$begingroup$
How do you make the transformation? Shoudn't the integer after transformation $x->logfrac1x$ become $int_0^infty{sinx}*e^{x(1-b)}dx$. I used the substitution $logfrac1x=u$ , $e^{-u}=x$ then $-frac1xdx=du$. Don't know where I'm wrong. Can you tell me please how to solve $int_0^infty{sinx}*e^{x(1+b)}dx$? And thanks for the help.
$endgroup$
– David Dan
Jan 11 at 18:30
add a comment |
$begingroup$
Using your definition, we see that there is a natural log in the denominator. An easy way to remove that is simply by using differentiation under the integral. Therefore, we differentiate with respect to $b$ because$$frac {mathrm dx^b}{mathrm db}=x^blog x$$Hence$$begin{align*}mathfrak I'(a,b) & =intlimits_0^1mathrm dx, x^bsinleft(logfrac 1xright)end{align*}$$Now make a transformation $xmapstologtfrac 1x$ to get rid of the nested sine - log function. Therefore, the limits change to zero and infinity, so that$$begin{align*}mathfrak{I}'(a,b) & =intlimits_0^{infty}mathrm dx,e^{x(1+b)}sin x\ & =frac 1{1+(1+b)^2}end{align*}$$Where the last line is derived by integrating the integral by parts twice. Since we have $mathfrak I'(a,b)$, we integrate it back with respect to $b$ to get$$mathfrak{I}(a,b)=arctan(1+b)+C$$To find $C$, we observe that when $a=b$, then $mathfrak{I}(a,b)=0$. Hence, we get that $C=arctan(1+a)$ so the final answer becomes$$intlimits_0^1mathrm dx,frac {x^b-x^a}{log x}sinleft(logfrac 1xright)color{blue}{=arctan(1+b)-arctan(1+a)}$$
answered Jan 11 at 16:16
Frank W.Frank W.
3,5731321
$endgroup$
$begingroup$
How do you make the transformation? Shoudn't the integer after transformation $x->logfrac1x$ become $int_0^infty{sinx}*e^{x(1-b)}dx$. I used the substitution $logfrac1x=u$ , $e^{-u}=x$ then $-frac1xdx=du$. Don't know where I'm wrong. Can you tell me please how to solve $int_0^infty{sinx}*e^{x(1+b)}dx$? And thanks for the help.
$endgroup$
– David Dan
Jan 11 at 18:30
add a comment |
$begingroup$
Using your definition, we see that there is a natural log in the denominator. An easy way to remove that is simply by using differentiation under the integral. Therefore, we differentiate with respect to $b$ because$$frac {mathrm dx^b}{mathrm db}=x^blog x$$Hence$$begin{align*}mathfrak I'(a,b) & =intlimits_0^1mathrm dx, x^bsinleft(logfrac 1xright)end{align*}$$Now make a transformation $xmapstologtfrac 1x$ to get rid of the nested sine - log function. Therefore, the limits change to zero and infinity, so that$$begin{align*}mathfrak{I}'(a,b) & =intlimits_0^{infty}mathrm dx,e^{x(1+b)}sin x\ & =frac 1{1+(1+b)^2}end{align*}$$Where the last line is derived by integrating the integral by parts twice. Since we have $mathfrak I'(a,b)$, we integrate it back with respect to $b$ to get$$mathfrak{I}(a,b)=arctan(1+b)+C$$To find $C$, we observe that when $a=b$, then $mathfrak{I}(a,b)=0$. Hence, we get that $C=arctan(1+a)$ so the final answer becomes$$intlimits_0^1mathrm dx,frac {x^b-x^a}{log x}sinleft(logfrac 1xright)color{blue}{=arctan(1+b)-arctan(1+a)}$$
answered Jan 11 at 16:16
Frank W.Frank W.
3,5731321
$endgroup$
Using your definition, we see that there is a natural log in the denominator. An easy way to remove that is simply by using differentiation under the integral. Therefore, we differentiate with respect to $b$ because$$frac {mathrm dx^b}{mathrm db}=x^blog x$$Hence$$begin{align*}mathfrak I'(a,b) & =intlimits_0^1mathrm dx, x^bsinleft(logfrac 1xright)end{align*}$$Now make a transformation $xmapstologtfrac 1x$ to get rid of the nested sine - log function. Therefore, the limits change to zero and infinity, so that$$begin{align*}mathfrak{I}'(a,b) & =intlimits_0^{infty}mathrm dx,e^{x(1+b)}sin x\ & =frac 1{1+(1+b)^2}end{align*}$$Where the last line is derived by integrating the integral by parts twice. Since we have $mathfrak I'(a,b)$, we integrate it back with respect to $b$ to get$$mathfrak{I}(a,b)=arctan(1+b)+C$$To find $C$, we observe that when $a=b$, then $mathfrak{I}(a,b)=0$. Hence, we get that $C=arctan(1+a)$ so the final answer becomes$$intlimits_0^1mathrm dx,frac {x^b-x^a}{log x}sinleft(logfrac 1xright)color{blue}{=arctan(1+b)-arctan(1+a)}$$
answered Jan 11 at 16:16
Frank W.Frank W.
3,5731321
answered Jan 11 at 16:16
Frank W.Frank W.
3,5731321
answered Jan 11 at 16:16
Frank W.Frank W.
3,5731321
answered Jan 11 at 16:16
Frank W.Frank W.
3,5731321
3,5731321
$begingroup$
How do you make the transformation? Shoudn't the integer after transformation $x->logfrac1x$ become $int_0^infty{sinx}*e^{x(1-b)}dx$. I used the substitution $logfrac1x=u$ , $e^{-u}=x$ then $-frac1xdx=du$. Don't know where I'm wrong. Can you tell me please how to solve $int_0^infty{sinx}*e^{x(1+b)}dx$? And thanks for the help.
$endgroup$
– David Dan
Jan 11 at 18:30
add a comment |
$begingroup$
How do you make the transformation? Shoudn't the integer after transformation $x->logfrac1x$ become $int_0^infty{sinx}*e^{x(1-b)}dx$. I used the substitution $logfrac1x=u$ , $e^{-u}=x$ then $-frac1xdx=du$. Don't know where I'm wrong. Can you tell me please how to solve $int_0^infty{sinx}*e^{x(1+b)}dx$? And thanks for the help.
$endgroup$
– David Dan
Jan 11 at 18:30
$begingroup$
How do you make the transformation? Shoudn't the integer after transformation $x->logfrac1x$ become $int_0^infty{sinx}*e^{x(1-b)}dx$. I used the substitution $logfrac1x=u$ , $e^{-u}=x$ then $-frac1xdx=du$. Don't know where I'm wrong. Can you tell me please how to solve $int_0^infty{sinx}*e^{x(1+b)}dx$? And thanks for the help.
$endgroup$
– David Dan
Jan 11 at 18:30
$begingroup$
How do you make the transformation? Shoudn't the integer after transformation $x->logfrac1x$ become $int_0^infty{sinx}*e^{x(1-b)}dx$. I used the substitution $logfrac1x=u$ , $e^{-u}=x$ then $-frac1xdx=du$. Don't know where I'm wrong. Can you tell me please how to solve $int_0^infty{sinx}*e^{x(1+b)}dx$? And thanks for the help.
$endgroup$
– David Dan
Jan 11 at 18:30
add a comment |
$begingroup$
First, under $xto e^{-x}$
$$ frac{partial I(a,b)}{partial b}=int_0^1sin(ln(frac1x))x^bdx=int_0^infty e^{(b+1)x}sin xdx=frac{1}{(b+1)^2+1} $$
and hence
$$ I(a,b)=int_a^bfrac{1}{(s+1)^2+1}ds=arctan (b+1)-arctan(a+1). $$
answered Jan 11 at 16:07
xpaulxpaul
22.7k24455
$endgroup$
add a comment |
$begingroup$
First, under $xto e^{-x}$
$$ frac{partial I(a,b)}{partial b}=int_0^1sin(ln(frac1x))x^bdx=int_0^infty e^{(b+1)x}sin xdx=frac{1}{(b+1)^2+1} $$
and hence
$$ I(a,b)=int_a^bfrac{1}{(s+1)^2+1}ds=arctan (b+1)-arctan(a+1). $$
answered Jan 11 at 16:07
xpaulxpaul
22.7k24455
$endgroup$
add a comment |
$begingroup$
First, under $xto e^{-x}$
$$ frac{partial I(a,b)}{partial b}=int_0^1sin(ln(frac1x))x^bdx=int_0^infty e^{(b+1)x}sin xdx=frac{1}{(b+1)^2+1} $$
and hence
$$ I(a,b)=int_a^bfrac{1}{(s+1)^2+1}ds=arctan (b+1)-arctan(a+1). $$
answered Jan 11 at 16:07
xpaulxpaul
22.7k24455
$endgroup$
First, under $xto e^{-x}$
$$ frac{partial I(a,b)}{partial b}=int_0^1sin(ln(frac1x))x^bdx=int_0^infty e^{(b+1)x}sin xdx=frac{1}{(b+1)^2+1} $$
and hence
$$ I(a,b)=int_a^bfrac{1}{(s+1)^2+1}ds=arctan (b+1)-arctan(a+1). $$
answered Jan 11 at 16:07
xpaulxpaul
22.7k24455
edited Jan 11 at 16:52
answered Jan 11 at 16:07
xpaulxpaul
22.7k24455
answered Jan 11 at 16:07
xpaulxpaul
22.7k24455
answered Jan 11 at 16:07
xpaulxpaul
22.7k24455
22.7k24455
add a comment |
add a comment |
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Required, but never shown
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the question is to calculate the integrals with parameters.I have found this problem in the proposed problems chapter at improper integrals.[drive.google.com/open?id=152R-HdfOJK7Y85YqQUZeVCGHmUOCvz2B] [drive.google.com/open?id=152R-HdfOJK7Y85YqQUZeVCGHmUOCvz2B] are the theorems i usually use for this kind of problem
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– David Dan
Jan 11 at 12:42
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new links in case the ones above doesn't work [imgur.com/ZL15gbO] imgur.com/gallery/MHVcQeE
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– David Dan
Jan 11 at 12:50
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Note that $$sin[log(1/x)]=sin(-log x)=-sinlog x$$
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– clathratus
Jan 11 at 16:57
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Then set $$J(s;a,b)=int_0^1sin(slog x)frac{x^a-x^b}{log x}dx$$ and note that $J(0;a,b)=0$ as well as $J(1;a,b)=I(a,b)$. Then take $frac{partial}{partial s}$ on both sides to see that $$frac{partial}{partial s}J(s;a,b)=int_0^1(x^a-x^b)cos(s log x)dx$$ Which may be easier
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– clathratus
Jan 11 at 17:03