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







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    4












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






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      share|cite|improve this question











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      asked yesterday









      Sher AfghanSher Afghan

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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.






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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).
            $endgroup$
            – ruakh
            yesterday











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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.






          share|cite|improve this answer









          $endgroup$









          • 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
















          8












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$









          • 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














          8












          8








          8





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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














          • 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


















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