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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    2












    $begingroup$


    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









    $endgroup$















      2












      2








      2





      $begingroup$


      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






      share|cite|improve this question













      share|cite|improve this question











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      asked 2 hours ago









      Sher AfghanSher Afghan

      1519




      1519






















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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.






          share|cite|improve this answer









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          • 1




            $begingroup$
            +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
            37 mins ago











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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









          $endgroup$









          • 1




            $begingroup$
            +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
            37 mins ago
















          6












          $begingroup$

          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









          $endgroup$









          • 1




            $begingroup$
            +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
            37 mins ago














          6












          6








          6





          $begingroup$

          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 1 hour ago









          herb steinbergherb steinberg

          2,6732310




          2,6732310








          • 1




            $begingroup$
            +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
            37 mins ago














          • 1




            $begingroup$
            +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
            37 mins ago








          1




          1




          $begingroup$
          +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
          37 mins ago




          $begingroup$
          +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
          37 mins ago


















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