Evaluating $int_{1}^inftyfrac{1}{t^asqrt{t^2-1}}dt$ for $ageq 1$












-1












$begingroup$


I know that the this integral converges, but I can't show it. And, how can I proceed to calculate its value?



$$displaystyle int_{1}^infty dfrac{1}{t^asqrt{t^2-1}} dtqquad (a geq 1)$$






improper-integrals







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  • $begingroup$
    i = 1 i'm sorry i edited it.
    $endgroup$
    – Suna
    Dec 6 '18 at 19:06










  • $begingroup$
    i tried computing its value using different values of a and for all $age 1 $ it always converged.
    $endgroup$
    – Suna
    Dec 6 '18 at 19:10










  • $begingroup$
    It converges when $displaystyleReleft(aright) > 0$.
    $endgroup$
    – Felix Marin
    Dec 6 '18 at 19:25












  • $begingroup$
    Did you try integration by parts ?
    $endgroup$
    – Damien
    Dec 6 '18 at 19:38


















-1












$begingroup$


I know that the this integral converges, but I can't show it. And, how can I proceed to calculate its value?



$$displaystyle int_{1}^infty dfrac{1}{t^asqrt{t^2-1}} dtqquad (a geq 1)$$






improper-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    i = 1 i'm sorry i edited it.
    $endgroup$
    – Suna
    Dec 6 '18 at 19:06










  • $begingroup$
    i tried computing its value using different values of a and for all $age 1 $ it always converged.
    $endgroup$
    – Suna
    Dec 6 '18 at 19:10










  • $begingroup$
    It converges when $displaystyleReleft(aright) > 0$.
    $endgroup$
    – Felix Marin
    Dec 6 '18 at 19:25












  • $begingroup$
    Did you try integration by parts ?
    $endgroup$
    – Damien
    Dec 6 '18 at 19:38
















-1












-1








-1





$begingroup$


I know that the this integral converges, but I can't show it. And, how can I proceed to calculate its value?



$$displaystyle int_{1}^infty dfrac{1}{t^asqrt{t^2-1}} dtqquad (a geq 1)$$






improper-integrals







share|cite|improve this question











$endgroup$




I know that the this integral converges, but I can't show it. And, how can I proceed to calculate its value?



$$displaystyle int_{1}^infty dfrac{1}{t^asqrt{t^2-1}} dtqquad (a geq 1)$$






improper-integrals




improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 19:41









Blue

47.8k870152




47.8k870152










asked Dec 6 '18 at 18:57









SunaSuna

33




33












  • $begingroup$
    i = 1 i'm sorry i edited it.
    $endgroup$
    – Suna
    Dec 6 '18 at 19:06










  • $begingroup$
    i tried computing its value using different values of a and for all $age 1 $ it always converged.
    $endgroup$
    – Suna
    Dec 6 '18 at 19:10










  • $begingroup$
    It converges when $displaystyleReleft(aright) > 0$.
    $endgroup$
    – Felix Marin
    Dec 6 '18 at 19:25












  • $begingroup$
    Did you try integration by parts ?
    $endgroup$
    – Damien
    Dec 6 '18 at 19:38




















  • $begingroup$
    i = 1 i'm sorry i edited it.
    $endgroup$
    – Suna
    Dec 6 '18 at 19:06










  • $begingroup$
    i tried computing its value using different values of a and for all $age 1 $ it always converged.
    $endgroup$
    – Suna
    Dec 6 '18 at 19:10










  • $begingroup$
    It converges when $displaystyleReleft(aright) > 0$.
    $endgroup$
    – Felix Marin
    Dec 6 '18 at 19:25












  • $begingroup$
    Did you try integration by parts ?
    $endgroup$
    – Damien
    Dec 6 '18 at 19:38


















$begingroup$
i = 1 i'm sorry i edited it.
$endgroup$
– Suna
Dec 6 '18 at 19:06




$begingroup$
i = 1 i'm sorry i edited it.
$endgroup$
– Suna
Dec 6 '18 at 19:06












$begingroup$
i tried computing its value using different values of a and for all $age 1 $ it always converged.
$endgroup$
– Suna
Dec 6 '18 at 19:10




$begingroup$
i tried computing its value using different values of a and for all $age 1 $ it always converged.
$endgroup$
– Suna
Dec 6 '18 at 19:10












$begingroup$
It converges when $displaystyleReleft(aright) > 0$.
$endgroup$
– Felix Marin
Dec 6 '18 at 19:25






$begingroup$
It converges when $displaystyleReleft(aright) > 0$.
$endgroup$
– Felix Marin
Dec 6 '18 at 19:25














$begingroup$
Did you try integration by parts ?
$endgroup$
– Damien
Dec 6 '18 at 19:38






$begingroup$
Did you try integration by parts ?
$endgroup$
– Damien
Dec 6 '18 at 19:38












2 Answers
2






active

oldest

votes


















1












$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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begin{align}
&bbox[10px,#ffd]{left.int_{1}^{infty}
{dd t over t^{a}root{t^{2} - 1}},rightvert
_{ Repars{a} > 0}}
,,,stackrel{t mapsto 1/t}{=},,,
int_{1}^{0}
{-dd t/t^{2} over pars{1/t}^{a}root{1/t^{2} - 1}}
\[5mm] = &
int_{0}^{1}t^{a - 1}pars{1 - t^{2}}^{-1/2},dd t
,,,stackrel{t^{2} mapsto t}{=},,,
{1 over 2}int_{0}^{1}t^{a/2 - 1}pars{1 - t}^{-1/2},dd t
\[5mm] = &
{1 over 2},{Gammapars{a/2}Gammapars{1/2} overGammapars{a/2 + 1/2}} =
bbx{{Gammapars{a/2} overGammapars{a/2 + 1/2}},
{root{pi} over 2}}
end{align}






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint:
    I would do the substitution $$t=frac1{cos x}$$
    When $t=1$, you have $x=0$, and for $t=infty$ you get $x=pi/2$. You also then have $sin xge 0$ in this interval, so $$sqrt{t^2-1}=frac{sin x}{cos x}=tan x$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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      newcommand{ds}[1]{displaystyle{#1}}
      newcommand{expo}[1]{,mathrm{e}^{#1},}
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      newcommand{mrm}[1]{mathrm{#1}}
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      newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
      newcommand{root}[2]{,sqrt[#1]{,{#2},},}
      newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
      newcommand{verts}[1]{leftvert,{#1},rightvert}$
      begin{align}
      &bbox[10px,#ffd]{left.int_{1}^{infty}
      {dd t over t^{a}root{t^{2} - 1}},rightvert
      _{ Repars{a} > 0}}
      ,,,stackrel{t mapsto 1/t}{=},,,
      int_{1}^{0}
      {-dd t/t^{2} over pars{1/t}^{a}root{1/t^{2} - 1}}
      \[5mm] = &
      int_{0}^{1}t^{a - 1}pars{1 - t^{2}}^{-1/2},dd t
      ,,,stackrel{t^{2} mapsto t}{=},,,
      {1 over 2}int_{0}^{1}t^{a/2 - 1}pars{1 - t}^{-1/2},dd t
      \[5mm] = &
      {1 over 2},{Gammapars{a/2}Gammapars{1/2} overGammapars{a/2 + 1/2}} =
      bbx{{Gammapars{a/2} overGammapars{a/2 + 1/2}},
      {root{pi} over 2}}
      end{align}






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
        newcommand{dd}{mathrm{d}}
        newcommand{ds}[1]{displaystyle{#1}}
        newcommand{expo}[1]{,mathrm{e}^{#1},}
        newcommand{ic}{mathrm{i}}
        newcommand{mc}[1]{mathcal{#1}}
        newcommand{mrm}[1]{mathrm{#1}}
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        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$
        begin{align}
        &bbox[10px,#ffd]{left.int_{1}^{infty}
        {dd t over t^{a}root{t^{2} - 1}},rightvert
        _{ Repars{a} > 0}}
        ,,,stackrel{t mapsto 1/t}{=},,,
        int_{1}^{0}
        {-dd t/t^{2} over pars{1/t}^{a}root{1/t^{2} - 1}}
        \[5mm] = &
        int_{0}^{1}t^{a - 1}pars{1 - t^{2}}^{-1/2},dd t
        ,,,stackrel{t^{2} mapsto t}{=},,,
        {1 over 2}int_{0}^{1}t^{a/2 - 1}pars{1 - t}^{-1/2},dd t
        \[5mm] = &
        {1 over 2},{Gammapars{a/2}Gammapars{1/2} overGammapars{a/2 + 1/2}} =
        bbx{{Gammapars{a/2} overGammapars{a/2 + 1/2}},
        {root{pi} over 2}}
        end{align}






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
          newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
          newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
          newcommand{dd}{mathrm{d}}
          newcommand{ds}[1]{displaystyle{#1}}
          newcommand{expo}[1]{,mathrm{e}^{#1},}
          newcommand{ic}{mathrm{i}}
          newcommand{mc}[1]{mathcal{#1}}
          newcommand{mrm}[1]{mathrm{#1}}
          newcommand{pars}[1]{left(,{#1},right)}
          newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
          newcommand{root}[2]{,sqrt[#1]{,{#2},},}
          newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
          newcommand{verts}[1]{leftvert,{#1},rightvert}$
          begin{align}
          &bbox[10px,#ffd]{left.int_{1}^{infty}
          {dd t over t^{a}root{t^{2} - 1}},rightvert
          _{ Repars{a} > 0}}
          ,,,stackrel{t mapsto 1/t}{=},,,
          int_{1}^{0}
          {-dd t/t^{2} over pars{1/t}^{a}root{1/t^{2} - 1}}
          \[5mm] = &
          int_{0}^{1}t^{a - 1}pars{1 - t^{2}}^{-1/2},dd t
          ,,,stackrel{t^{2} mapsto t}{=},,,
          {1 over 2}int_{0}^{1}t^{a/2 - 1}pars{1 - t}^{-1/2},dd t
          \[5mm] = &
          {1 over 2},{Gammapars{a/2}Gammapars{1/2} overGammapars{a/2 + 1/2}} =
          bbx{{Gammapars{a/2} overGammapars{a/2 + 1/2}},
          {root{pi} over 2}}
          end{align}






          share|cite|improve this answer









          $endgroup$



          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
          newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
          newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
          newcommand{dd}{mathrm{d}}
          newcommand{ds}[1]{displaystyle{#1}}
          newcommand{expo}[1]{,mathrm{e}^{#1},}
          newcommand{ic}{mathrm{i}}
          newcommand{mc}[1]{mathcal{#1}}
          newcommand{mrm}[1]{mathrm{#1}}
          newcommand{pars}[1]{left(,{#1},right)}
          newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
          newcommand{root}[2]{,sqrt[#1]{,{#2},},}
          newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
          newcommand{verts}[1]{leftvert,{#1},rightvert}$
          begin{align}
          &bbox[10px,#ffd]{left.int_{1}^{infty}
          {dd t over t^{a}root{t^{2} - 1}},rightvert
          _{ Repars{a} > 0}}
          ,,,stackrel{t mapsto 1/t}{=},,,
          int_{1}^{0}
          {-dd t/t^{2} over pars{1/t}^{a}root{1/t^{2} - 1}}
          \[5mm] = &
          int_{0}^{1}t^{a - 1}pars{1 - t^{2}}^{-1/2},dd t
          ,,,stackrel{t^{2} mapsto t}{=},,,
          {1 over 2}int_{0}^{1}t^{a/2 - 1}pars{1 - t}^{-1/2},dd t
          \[5mm] = &
          {1 over 2},{Gammapars{a/2}Gammapars{1/2} overGammapars{a/2 + 1/2}} =
          bbx{{Gammapars{a/2} overGammapars{a/2 + 1/2}},
          {root{pi} over 2}}
          end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 19:40









          Felix MarinFelix Marin

          67.4k7107141




          67.4k7107141























              0












              $begingroup$

              Hint:
              I would do the substitution $$t=frac1{cos x}$$
              When $t=1$, you have $x=0$, and for $t=infty$ you get $x=pi/2$. You also then have $sin xge 0$ in this interval, so $$sqrt{t^2-1}=frac{sin x}{cos x}=tan x$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint:
                I would do the substitution $$t=frac1{cos x}$$
                When $t=1$, you have $x=0$, and for $t=infty$ you get $x=pi/2$. You also then have $sin xge 0$ in this interval, so $$sqrt{t^2-1}=frac{sin x}{cos x}=tan x$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint:
                  I would do the substitution $$t=frac1{cos x}$$
                  When $t=1$, you have $x=0$, and for $t=infty$ you get $x=pi/2$. You also then have $sin xge 0$ in this interval, so $$sqrt{t^2-1}=frac{sin x}{cos x}=tan x$$






                  share|cite|improve this answer









                  $endgroup$



                  Hint:
                  I would do the substitution $$t=frac1{cos x}$$
                  When $t=1$, you have $x=0$, and for $t=infty$ you get $x=pi/2$. You also then have $sin xge 0$ in this interval, so $$sqrt{t^2-1}=frac{sin x}{cos x}=tan x$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 19:36









                  AndreiAndrei

                  11.7k21026




                  11.7k21026






























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