How do I show that $F(1/x) = -F(x)$?












1














Let $F$ be defined by



$F(x) = int_1^x frac{exp(frac{u^2 + 1}{u})du}{u}$.



I'm supposed to show that $F(1/x) = -F(x)$. The hint in the back of my book says that I should use the substitution $v = frac{1}{u}$. When I do this I get the integral



$int_1^x frac{-1}{v^2}exp(frac{1}{v} + v)dv$. Am I supposed to proceed by doing integration by parts?






calculus analysis







share|cite|improve this question




















  • 1




    You need to change your integration limits after making the substitution.
    – Mattos
    Nov 29 '18 at 22:37
















1














Let $F$ be defined by



$F(x) = int_1^x frac{exp(frac{u^2 + 1}{u})du}{u}$.



I'm supposed to show that $F(1/x) = -F(x)$. The hint in the back of my book says that I should use the substitution $v = frac{1}{u}$. When I do this I get the integral



$int_1^x frac{-1}{v^2}exp(frac{1}{v} + v)dv$. Am I supposed to proceed by doing integration by parts?






calculus analysis







share|cite|improve this question




















  • 1




    You need to change your integration limits after making the substitution.
    – Mattos
    Nov 29 '18 at 22:37














1












1








1







Let $F$ be defined by



$F(x) = int_1^x frac{exp(frac{u^2 + 1}{u})du}{u}$.



I'm supposed to show that $F(1/x) = -F(x)$. The hint in the back of my book says that I should use the substitution $v = frac{1}{u}$. When I do this I get the integral



$int_1^x frac{-1}{v^2}exp(frac{1}{v} + v)dv$. Am I supposed to proceed by doing integration by parts?






calculus analysis







share|cite|improve this question















Let $F$ be defined by



$F(x) = int_1^x frac{exp(frac{u^2 + 1}{u})du}{u}$.



I'm supposed to show that $F(1/x) = -F(x)$. The hint in the back of my book says that I should use the substitution $v = frac{1}{u}$. When I do this I get the integral



$int_1^x frac{-1}{v^2}exp(frac{1}{v} + v)dv$. Am I supposed to proceed by doing integration by parts?






calculus analysis




calculus analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 22:39









Bernard

118k639112




118k639112










asked Nov 29 '18 at 22:33









K.MK.M

686312




686312








  • 1




    You need to change your integration limits after making the substitution.
    – Mattos
    Nov 29 '18 at 22:37














  • 1




    You need to change your integration limits after making the substitution.
    – Mattos
    Nov 29 '18 at 22:37








1




1




You need to change your integration limits after making the substitution.
– Mattos
Nov 29 '18 at 22:37




You need to change your integration limits after making the substitution.
– Mattos
Nov 29 '18 at 22:37










2 Answers
2






active

oldest

votes


















2














The definition of $F(x)$ is
$$F(x)=int_1^x e^{u+1/u}frac{du}{u}$$
Substituting $umapsto 1/v$ sends $e^{u+1/u}mapsto e^{v+1/v}$ and $du/umapsto-dv/v$, and replaces your integration bounds with $1$ and $1/x$ respectively, so you have
$$F(x)=int_1^{1/x} e^{v+1/v}cdot -frac{dv}{v}=-F(1/x)$$
and you are done.






share|cite|improve this answer





















  • I just want to make sure that the order of the bounds doesn't matter. So even if $1/x < 1$, we can still write the integral as you've done above?
    – K.M
    Nov 29 '18 at 23:10










  • @K.M Yes! Recall the following property of integrals: $$int_a^b=-int_b^a$$
    – Frpzzd
    Nov 29 '18 at 23:48



















2














If you do $v=dfrac1u$ and $mathrm dv=-dfrac{mathrm du}{u^2}$, then your integral becomes$$-int_1^{frac1x}frac{expleft(frac1v+vright)}{frac1v}timesfrac1{v^2},mathrm dv=-int_1^{frac1x}frac{expleft(frac1v+vright)}v,mathrm dv=-Fleft(frac1xright).$$






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    The definition of $F(x)$ is
    $$F(x)=int_1^x e^{u+1/u}frac{du}{u}$$
    Substituting $umapsto 1/v$ sends $e^{u+1/u}mapsto e^{v+1/v}$ and $du/umapsto-dv/v$, and replaces your integration bounds with $1$ and $1/x$ respectively, so you have
    $$F(x)=int_1^{1/x} e^{v+1/v}cdot -frac{dv}{v}=-F(1/x)$$
    and you are done.






    share|cite|improve this answer





















    • I just want to make sure that the order of the bounds doesn't matter. So even if $1/x < 1$, we can still write the integral as you've done above?
      – K.M
      Nov 29 '18 at 23:10










    • @K.M Yes! Recall the following property of integrals: $$int_a^b=-int_b^a$$
      – Frpzzd
      Nov 29 '18 at 23:48
















    2














    The definition of $F(x)$ is
    $$F(x)=int_1^x e^{u+1/u}frac{du}{u}$$
    Substituting $umapsto 1/v$ sends $e^{u+1/u}mapsto e^{v+1/v}$ and $du/umapsto-dv/v$, and replaces your integration bounds with $1$ and $1/x$ respectively, so you have
    $$F(x)=int_1^{1/x} e^{v+1/v}cdot -frac{dv}{v}=-F(1/x)$$
    and you are done.






    share|cite|improve this answer





















    • I just want to make sure that the order of the bounds doesn't matter. So even if $1/x < 1$, we can still write the integral as you've done above?
      – K.M
      Nov 29 '18 at 23:10










    • @K.M Yes! Recall the following property of integrals: $$int_a^b=-int_b^a$$
      – Frpzzd
      Nov 29 '18 at 23:48














    2












    2








    2






    The definition of $F(x)$ is
    $$F(x)=int_1^x e^{u+1/u}frac{du}{u}$$
    Substituting $umapsto 1/v$ sends $e^{u+1/u}mapsto e^{v+1/v}$ and $du/umapsto-dv/v$, and replaces your integration bounds with $1$ and $1/x$ respectively, so you have
    $$F(x)=int_1^{1/x} e^{v+1/v}cdot -frac{dv}{v}=-F(1/x)$$
    and you are done.






    share|cite|improve this answer












    The definition of $F(x)$ is
    $$F(x)=int_1^x e^{u+1/u}frac{du}{u}$$
    Substituting $umapsto 1/v$ sends $e^{u+1/u}mapsto e^{v+1/v}$ and $du/umapsto-dv/v$, and replaces your integration bounds with $1$ and $1/x$ respectively, so you have
    $$F(x)=int_1^{1/x} e^{v+1/v}cdot -frac{dv}{v}=-F(1/x)$$
    and you are done.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 29 '18 at 22:39









    FrpzzdFrpzzd

    22.3k839107




    22.3k839107












    • I just want to make sure that the order of the bounds doesn't matter. So even if $1/x < 1$, we can still write the integral as you've done above?
      – K.M
      Nov 29 '18 at 23:10










    • @K.M Yes! Recall the following property of integrals: $$int_a^b=-int_b^a$$
      – Frpzzd
      Nov 29 '18 at 23:48


















    • I just want to make sure that the order of the bounds doesn't matter. So even if $1/x < 1$, we can still write the integral as you've done above?
      – K.M
      Nov 29 '18 at 23:10










    • @K.M Yes! Recall the following property of integrals: $$int_a^b=-int_b^a$$
      – Frpzzd
      Nov 29 '18 at 23:48
















    I just want to make sure that the order of the bounds doesn't matter. So even if $1/x < 1$, we can still write the integral as you've done above?
    – K.M
    Nov 29 '18 at 23:10




    I just want to make sure that the order of the bounds doesn't matter. So even if $1/x < 1$, we can still write the integral as you've done above?
    – K.M
    Nov 29 '18 at 23:10












    @K.M Yes! Recall the following property of integrals: $$int_a^b=-int_b^a$$
    – Frpzzd
    Nov 29 '18 at 23:48




    @K.M Yes! Recall the following property of integrals: $$int_a^b=-int_b^a$$
    – Frpzzd
    Nov 29 '18 at 23:48











    2














    If you do $v=dfrac1u$ and $mathrm dv=-dfrac{mathrm du}{u^2}$, then your integral becomes$$-int_1^{frac1x}frac{expleft(frac1v+vright)}{frac1v}timesfrac1{v^2},mathrm dv=-int_1^{frac1x}frac{expleft(frac1v+vright)}v,mathrm dv=-Fleft(frac1xright).$$






    share|cite|improve this answer


























      2














      If you do $v=dfrac1u$ and $mathrm dv=-dfrac{mathrm du}{u^2}$, then your integral becomes$$-int_1^{frac1x}frac{expleft(frac1v+vright)}{frac1v}timesfrac1{v^2},mathrm dv=-int_1^{frac1x}frac{expleft(frac1v+vright)}v,mathrm dv=-Fleft(frac1xright).$$






      share|cite|improve this answer
























        2












        2








        2






        If you do $v=dfrac1u$ and $mathrm dv=-dfrac{mathrm du}{u^2}$, then your integral becomes$$-int_1^{frac1x}frac{expleft(frac1v+vright)}{frac1v}timesfrac1{v^2},mathrm dv=-int_1^{frac1x}frac{expleft(frac1v+vright)}v,mathrm dv=-Fleft(frac1xright).$$






        share|cite|improve this answer












        If you do $v=dfrac1u$ and $mathrm dv=-dfrac{mathrm du}{u^2}$, then your integral becomes$$-int_1^{frac1x}frac{expleft(frac1v+vright)}{frac1v}timesfrac1{v^2},mathrm dv=-int_1^{frac1x}frac{expleft(frac1v+vright)}v,mathrm dv=-Fleft(frac1xright).$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 22:39









        José Carlos SantosJosé Carlos Santos

        152k22123225




        152k22123225






























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