The integral of a function multiplied with an unbounded function converges if the product is a bounded...
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Suppose $f:[1,infty]rightarrow mathbb{R}$ is such that $g(x):=x^2f(x)$ is a bounded function. Prove that $int_1^infty f$ converges.
Intuitively I think it seems reasonable. As $x^2$ is increasing and unbounded above, $f(x)$ must be a decreasing function with higher absolute slope than $x^2$ so as $g(x)$ can be bounded. Not sure how to proceed further from here.
real-analysis convergence improper-integrals
asked yesterday
Sher AfghanSher Afghan
1619
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Suppose $f:[1,infty]rightarrow mathbb{R}$ is such that $g(x):=x^2f(x)$ is a bounded function. Prove that $int_1^infty f$ converges.
Intuitively I think it seems reasonable. As $x^2$ is increasing and unbounded above, $f(x)$ must be a decreasing function with higher absolute slope than $x^2$ so as $g(x)$ can be bounded. Not sure how to proceed further from here.
real-analysis convergence improper-integrals
asked yesterday
Sher AfghanSher Afghan
1619
$endgroup$
add a comment |
$begingroup$
Suppose $f:[1,infty]rightarrow mathbb{R}$ is such that $g(x):=x^2f(x)$ is a bounded function. Prove that $int_1^infty f$ converges.
Intuitively I think it seems reasonable. As $x^2$ is increasing and unbounded above, $f(x)$ must be a decreasing function with higher absolute slope than $x^2$ so as $g(x)$ can be bounded. Not sure how to proceed further from here.
real-analysis convergence improper-integrals
asked yesterday
Sher AfghanSher Afghan
1619
$endgroup$
Suppose $f:[1,infty]rightarrow mathbb{R}$ is such that $g(x):=x^2f(x)$ is a bounded function. Prove that $int_1^infty f$ converges.
Intuitively I think it seems reasonable. As $x^2$ is increasing and unbounded above, $f(x)$ must be a decreasing function with higher absolute slope than $x^2$ so as $g(x)$ can be bounded. Not sure how to proceed further from here.
real-analysis convergence improper-integrals
real-analysis convergence improper-integrals
asked yesterday
Sher AfghanSher Afghan
1619
asked yesterday
Sher AfghanSher Afghan
1619
asked yesterday
Sher AfghanSher Afghan
1619
asked yesterday
Sher AfghanSher Afghan
1619
asked yesterday
Sher AfghanSher Afghan
1619
1619
add a comment |
add a comment |
1 Answer
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Let $|g(x)|le M$ (assumption it is bounded). Then $|f(x)|le frac{M}{x^2}$ so $int_1^infty|f(x)|dxle Mint_1^inftyfrac{dx}{x^2}=M$. Therefore $f(x)$ is integrable.
answered yesterday
herb steinbergherb steinberg
2,6932310
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4
$begingroup$
+1, but just to make it explicit: the OP's intuitive reasoning about "increasing and unbounded above" is not correct, because the claim becomes false if we replace $g(x) := x^2 f(x)$ with $g(x) := x f(x)$ (even though $x$ is "increasing and unbounded above" just as $x^2$ is).
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– ruakh
yesterday
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Let $|g(x)|le M$ (assumption it is bounded). Then $|f(x)|le frac{M}{x^2}$ so $int_1^infty|f(x)|dxle Mint_1^inftyfrac{dx}{x^2}=M$. Therefore $f(x)$ is integrable.
answered yesterday
herb steinbergherb steinberg
2,6932310
$endgroup$
4
$begingroup$
+1, but just to make it explicit: the OP's intuitive reasoning about "increasing and unbounded above" is not correct, because the claim becomes false if we replace $g(x) := x^2 f(x)$ with $g(x) := x f(x)$ (even though $x$ is "increasing and unbounded above" just as $x^2$ is).
$endgroup$
– ruakh
yesterday
add a comment |
$begingroup$
Let $|g(x)|le M$ (assumption it is bounded). Then $|f(x)|le frac{M}{x^2}$ so $int_1^infty|f(x)|dxle Mint_1^inftyfrac{dx}{x^2}=M$. Therefore $f(x)$ is integrable.
answered yesterday
herb steinbergherb steinberg
2,6932310
$endgroup$
4
$begingroup$
+1, but just to make it explicit: the OP's intuitive reasoning about "increasing and unbounded above" is not correct, because the claim becomes false if we replace $g(x) := x^2 f(x)$ with $g(x) := x f(x)$ (even though $x$ is "increasing and unbounded above" just as $x^2$ is).
$endgroup$
– ruakh
yesterday
add a comment |
$begingroup$
Let $|g(x)|le M$ (assumption it is bounded). Then $|f(x)|le frac{M}{x^2}$ so $int_1^infty|f(x)|dxle Mint_1^inftyfrac{dx}{x^2}=M$. Therefore $f(x)$ is integrable.
answered yesterday
herb steinbergherb steinberg
2,6932310
$endgroup$
Let $|g(x)|le M$ (assumption it is bounded). Then $|f(x)|le frac{M}{x^2}$ so $int_1^infty|f(x)|dxle Mint_1^inftyfrac{dx}{x^2}=M$. Therefore $f(x)$ is integrable.
answered yesterday
herb steinbergherb steinberg
2,6932310
answered yesterday
herb steinbergherb steinberg
2,6932310
answered yesterday
herb steinbergherb steinberg
2,6932310
answered yesterday
herb steinbergherb steinberg
2,6932310
2,6932310
4
$begingroup$
+1, but just to make it explicit: the OP's intuitive reasoning about "increasing and unbounded above" is not correct, because the claim becomes false if we replace $g(x) := x^2 f(x)$ with $g(x) := x f(x)$ (even though $x$ is "increasing and unbounded above" just as $x^2$ is).
$endgroup$
– ruakh
yesterday
add a comment |
4
$begingroup$
+1, but just to make it explicit: the OP's intuitive reasoning about "increasing and unbounded above" is not correct, because the claim becomes false if we replace $g(x) := x^2 f(x)$ with $g(x) := x f(x)$ (even though $x$ is "increasing and unbounded above" just as $x^2$ is).
$endgroup$
– ruakh
yesterday
4
4
$begingroup$
+1, but just to make it explicit: the OP's intuitive reasoning about "increasing and unbounded above" is not correct, because the claim becomes false if we replace $g(x) := x^2 f(x)$ with $g(x) := x f(x)$ (even though $x$ is "increasing and unbounded above" just as $x^2$ is).
$endgroup$
– ruakh
yesterday
$begingroup$
+1, but just to make it explicit: the OP's intuitive reasoning about "increasing and unbounded above" is not correct, because the claim becomes false if we replace $g(x) := x^2 f(x)$ with $g(x) := x f(x)$ (even though $x$ is "increasing and unbounded above" just as $x^2$ is).
$endgroup$
– ruakh
yesterday
add a comment |
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