$int_0^{infty}frac{sin x}{(1+x)^2},dx$ converges absolutely
This is part of problem from baby Rudin Ch 6, exer 9:
How to show that $$int_0^{infty}frac{|sin x|}{(1+x)^2},dx$$ exists (i.e., $limlimits_{atoinfty}(int_0^{a}frac{|sin x|}{(1+x)^2},dx)$ exists)? (I know that $int_0^{infty}frac{sin x}{(1+x)^2},dx$ exists.)
I used integration calculator, and it suggested me to take note of $|x|=frac{x^2}{|x|}$, ie, they $$int_0^{infty}frac{|sin x|}{(1+x)^2},dx=int_0^{infty}frac{sin x}{|sin x|}frac{sin x}{(1+x)^2},dx,$$ but then they took $frac{sin x}{|sin x|}$ out, i.e., $$int_0^{infty}frac{sin x}{|sin x|}frac{sin x}{(1+x)^2},dx=frac{sin x}{|sin x|}int_0^{infty}frac{sin x}{(1+x)^2},dx,$$ which seems to me illegal.
real-analysis integration limits
asked Dec 12 '18 at 6:33
SilentSilent
2,67532150
add a comment |
This is part of problem from baby Rudin Ch 6, exer 9:
How to show that $$int_0^{infty}frac{|sin x|}{(1+x)^2},dx$$ exists (i.e., $limlimits_{atoinfty}(int_0^{a}frac{|sin x|}{(1+x)^2},dx)$ exists)? (I know that $int_0^{infty}frac{sin x}{(1+x)^2},dx$ exists.)
I used integration calculator, and it suggested me to take note of $|x|=frac{x^2}{|x|}$, ie, they $$int_0^{infty}frac{|sin x|}{(1+x)^2},dx=int_0^{infty}frac{sin x}{|sin x|}frac{sin x}{(1+x)^2},dx,$$ but then they took $frac{sin x}{|sin x|}$ out, i.e., $$int_0^{infty}frac{sin x}{|sin x|}frac{sin x}{(1+x)^2},dx=frac{sin x}{|sin x|}int_0^{infty}frac{sin x}{(1+x)^2},dx,$$ which seems to me illegal.
real-analysis integration limits
asked Dec 12 '18 at 6:33
SilentSilent
2,67532150
add a comment |
This is part of problem from baby Rudin Ch 6, exer 9:
How to show that $$int_0^{infty}frac{|sin x|}{(1+x)^2},dx$$ exists (i.e., $limlimits_{atoinfty}(int_0^{a}frac{|sin x|}{(1+x)^2},dx)$ exists)? (I know that $int_0^{infty}frac{sin x}{(1+x)^2},dx$ exists.)
I used integration calculator, and it suggested me to take note of $|x|=frac{x^2}{|x|}$, ie, they $$int_0^{infty}frac{|sin x|}{(1+x)^2},dx=int_0^{infty}frac{sin x}{|sin x|}frac{sin x}{(1+x)^2},dx,$$ but then they took $frac{sin x}{|sin x|}$ out, i.e., $$int_0^{infty}frac{sin x}{|sin x|}frac{sin x}{(1+x)^2},dx=frac{sin x}{|sin x|}int_0^{infty}frac{sin x}{(1+x)^2},dx,$$ which seems to me illegal.
real-analysis integration limits
asked Dec 12 '18 at 6:33
SilentSilent
2,67532150
This is part of problem from baby Rudin Ch 6, exer 9:
How to show that $$int_0^{infty}frac{|sin x|}{(1+x)^2},dx$$ exists (i.e., $limlimits_{atoinfty}(int_0^{a}frac{|sin x|}{(1+x)^2},dx)$ exists)? (I know that $int_0^{infty}frac{sin x}{(1+x)^2},dx$ exists.)
I used integration calculator, and it suggested me to take note of $|x|=frac{x^2}{|x|}$, ie, they $$int_0^{infty}frac{|sin x|}{(1+x)^2},dx=int_0^{infty}frac{sin x}{|sin x|}frac{sin x}{(1+x)^2},dx,$$ but then they took $frac{sin x}{|sin x|}$ out, i.e., $$int_0^{infty}frac{sin x}{|sin x|}frac{sin x}{(1+x)^2},dx=frac{sin x}{|sin x|}int_0^{infty}frac{sin x}{(1+x)^2},dx,$$ which seems to me illegal.
real-analysis integration limits
real-analysis integration limits
asked Dec 12 '18 at 6:33
SilentSilent
2,67532150
asked Dec 12 '18 at 6:33
SilentSilent
2,67532150
asked Dec 12 '18 at 6:33
SilentSilent
2,67532150
asked Dec 12 '18 at 6:33
SilentSilent
2,67532150
asked Dec 12 '18 at 6:33
SilentSilent
2,67532150
2,67532150
add a comment |
add a comment |
1 Answer
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Does
$$
int_0^{+infty} frac {vert sin x vert } {(1+x)^2} ,mathrm dx leqslant int_0^{+infty} frac {mathrm dx}{(1+x )^2}
$$
show the existence?
answered Dec 12 '18 at 6:37
xbhxbh
5,7651522
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Does
$$
int_0^{+infty} frac {vert sin x vert } {(1+x)^2} ,mathrm dx leqslant int_0^{+infty} frac {mathrm dx}{(1+x )^2}
$$
show the existence?
answered Dec 12 '18 at 6:37
xbhxbh
5,7651522
add a comment |
Does
$$
int_0^{+infty} frac {vert sin x vert } {(1+x)^2} ,mathrm dx leqslant int_0^{+infty} frac {mathrm dx}{(1+x )^2}
$$
show the existence?
answered Dec 12 '18 at 6:37
xbhxbh
5,7651522
add a comment |
Does
$$
int_0^{+infty} frac {vert sin x vert } {(1+x)^2} ,mathrm dx leqslant int_0^{+infty} frac {mathrm dx}{(1+x )^2}
$$
show the existence?
answered Dec 12 '18 at 6:37
xbhxbh
5,7651522
Does
$$
int_0^{+infty} frac {vert sin x vert } {(1+x)^2} ,mathrm dx leqslant int_0^{+infty} frac {mathrm dx}{(1+x )^2}
$$
show the existence?
answered Dec 12 '18 at 6:37
xbhxbh
5,7651522
answered Dec 12 '18 at 6:37
xbhxbh
5,7651522
answered Dec 12 '18 at 6:37
xbhxbh
5,7651522
answered Dec 12 '18 at 6:37
xbhxbh
5,7651522
5,7651522
add a comment |
add a comment |
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