Infinite Integration using the ceiling function











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$$int_0^infty frac{sin(xpi)}{lceil x rceil^2 + lceil x rceil} dx$$



My teacher recently gave me this and it's stumped me.






integration







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  • Do we have to find a closed form, or just study the convergence? The latter is immediate
    – Federico
    Dec 4 at 20:14










  • He just said find the value of that.
    – user546944
    Dec 4 at 20:17






  • 2




    I would suggest to take things slowly. Starting with $lim_{nto infty} int_0^n frac{1}{lceil x^2 rceil + lceil x rceil} dx$ then split up the integral into n pieces so that we get rid of the ceil functions.
    – Zacky
    Dec 4 at 20:19












  • start dividing the integral in pieces where $frac1{lceil x^2rceil +lceil xrceil}$ is constant
    – Masacroso
    Dec 4 at 20:25












  • This is unpleasant, since it depends on the interleaving between the values of $n$ and $sqrt{m}$ for $n,minmathbb{N}$. Are you sure it is $lceil x^2rceil$ and not $lceil x rceil^2$?
    – Jack D'Aurizio
    Dec 4 at 20:43

















up vote
5
down vote

favorite
5












$$int_0^infty frac{sin(xpi)}{lceil x rceil^2 + lceil x rceil} dx$$



My teacher recently gave me this and it's stumped me.






integration







share|cite|improve this question
























  • Do we have to find a closed form, or just study the convergence? The latter is immediate
    – Federico
    Dec 4 at 20:14










  • He just said find the value of that.
    – user546944
    Dec 4 at 20:17






  • 2




    I would suggest to take things slowly. Starting with $lim_{nto infty} int_0^n frac{1}{lceil x^2 rceil + lceil x rceil} dx$ then split up the integral into n pieces so that we get rid of the ceil functions.
    – Zacky
    Dec 4 at 20:19












  • start dividing the integral in pieces where $frac1{lceil x^2rceil +lceil xrceil}$ is constant
    – Masacroso
    Dec 4 at 20:25












  • This is unpleasant, since it depends on the interleaving between the values of $n$ and $sqrt{m}$ for $n,minmathbb{N}$. Are you sure it is $lceil x^2rceil$ and not $lceil x rceil^2$?
    – Jack D'Aurizio
    Dec 4 at 20:43















up vote
5
down vote

favorite
5









up vote
5
down vote

favorite
5






5





$$int_0^infty frac{sin(xpi)}{lceil x rceil^2 + lceil x rceil} dx$$



My teacher recently gave me this and it's stumped me.






integration







share|cite|improve this question















$$int_0^infty frac{sin(xpi)}{lceil x rceil^2 + lceil x rceil} dx$$



My teacher recently gave me this and it's stumped me.






integration




integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 at 21:23

























asked Dec 4 at 20:07







user546944



















  • Do we have to find a closed form, or just study the convergence? The latter is immediate
    – Federico
    Dec 4 at 20:14










  • He just said find the value of that.
    – user546944
    Dec 4 at 20:17






  • 2




    I would suggest to take things slowly. Starting with $lim_{nto infty} int_0^n frac{1}{lceil x^2 rceil + lceil x rceil} dx$ then split up the integral into n pieces so that we get rid of the ceil functions.
    – Zacky
    Dec 4 at 20:19












  • start dividing the integral in pieces where $frac1{lceil x^2rceil +lceil xrceil}$ is constant
    – Masacroso
    Dec 4 at 20:25












  • This is unpleasant, since it depends on the interleaving between the values of $n$ and $sqrt{m}$ for $n,minmathbb{N}$. Are you sure it is $lceil x^2rceil$ and not $lceil x rceil^2$?
    – Jack D'Aurizio
    Dec 4 at 20:43




















  • Do we have to find a closed form, or just study the convergence? The latter is immediate
    – Federico
    Dec 4 at 20:14










  • He just said find the value of that.
    – user546944
    Dec 4 at 20:17






  • 2




    I would suggest to take things slowly. Starting with $lim_{nto infty} int_0^n frac{1}{lceil x^2 rceil + lceil x rceil} dx$ then split up the integral into n pieces so that we get rid of the ceil functions.
    – Zacky
    Dec 4 at 20:19












  • start dividing the integral in pieces where $frac1{lceil x^2rceil +lceil xrceil}$ is constant
    – Masacroso
    Dec 4 at 20:25












  • This is unpleasant, since it depends on the interleaving between the values of $n$ and $sqrt{m}$ for $n,minmathbb{N}$. Are you sure it is $lceil x^2rceil$ and not $lceil x rceil^2$?
    – Jack D'Aurizio
    Dec 4 at 20:43


















Do we have to find a closed form, or just study the convergence? The latter is immediate
– Federico
Dec 4 at 20:14




Do we have to find a closed form, or just study the convergence? The latter is immediate
– Federico
Dec 4 at 20:14












He just said find the value of that.
– user546944
Dec 4 at 20:17




He just said find the value of that.
– user546944
Dec 4 at 20:17




2




2




I would suggest to take things slowly. Starting with $lim_{nto infty} int_0^n frac{1}{lceil x^2 rceil + lceil x rceil} dx$ then split up the integral into n pieces so that we get rid of the ceil functions.
– Zacky
Dec 4 at 20:19






I would suggest to take things slowly. Starting with $lim_{nto infty} int_0^n frac{1}{lceil x^2 rceil + lceil x rceil} dx$ then split up the integral into n pieces so that we get rid of the ceil functions.
– Zacky
Dec 4 at 20:19














start dividing the integral in pieces where $frac1{lceil x^2rceil +lceil xrceil}$ is constant
– Masacroso
Dec 4 at 20:25






start dividing the integral in pieces where $frac1{lceil x^2rceil +lceil xrceil}$ is constant
– Masacroso
Dec 4 at 20:25














This is unpleasant, since it depends on the interleaving between the values of $n$ and $sqrt{m}$ for $n,minmathbb{N}$. Are you sure it is $lceil x^2rceil$ and not $lceil x rceil^2$?
– Jack D'Aurizio
Dec 4 at 20:43






This is unpleasant, since it depends on the interleaving between the values of $n$ and $sqrt{m}$ for $n,minmathbb{N}$. Are you sure it is $lceil x^2rceil$ and not $lceil x rceil^2$?
– Jack D'Aurizio
Dec 4 at 20:43












2 Answers
2






active

oldest

votes

















up vote
8
down vote













$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}{sinpars{pi x} over
leftlceil{x}rightrceil^{2} + leftlceil{x}rightrceil}
,dd x} =
sum_{k = 0}^{infty}int_{k}^{k + 1}
{sinpars{pi x} over
pars{k + 1}^{2} + pars{k + 1}},dd x
\[5mm] = &
{2 over pi}sum_{k = 0}^{infty}
{pars{-1}^{k} over pars{k + 1}pars{k + 2}} =
{2 over pi}bracks{%
sum_{k = 0}^{infty}{pars{-1}^{k} over k + 1} -
sum_{k = 0}^{infty}{pars{-1}^{k} over k + 2}}
\[5mm] = &
{2 over pi}bracks{%
-sum_{k = 1}^{infty}{pars{-1}^{k} over k} -
sum_{k = 2}^{infty}{pars{-1}^{k} over k}}
\[5mm] = &
-,{2 over pi}braces{%
sum_{k = 1}^{infty}{pars{-1}^{k} over k} +
bracks{1 + sum_{k = 1}^{infty}{pars{-1}^{k} over k}}} \[5mm] = &
-,{2 over pi}bracks{%
1 + 2sum_{k = 1}^{infty}{pars{-1}^{k} over k}} =
bbx{4lnpars{2} - 2 over pi} approx 0.2459
end{align}






share|cite|improve this answer






























    up vote
    3
    down vote













    Hint: break the integral up into
    $$int_0^infty=int_0^1+int_1^2+int_2^3+cdots.$$
    On each of these intervals the denominator takes a constant value, so you can bring it out of the integral sign and evaluate each. After that just sum all the terms you get.






    share|cite|improve this answer

















    • 1




      Can the downvoter kindly explain?
      – YiFan
      Dec 5 at 2:07











    Your Answer





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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    8
    down vote













    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,mathrm{e}^{#1},}
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    newcommand{mc}[1]{mathcal{#1}}
    newcommand{mrm}[1]{mathrm{#1}}
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    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$
    begin{align}
    &bbox[10px,#ffd]{int_{0}^{infty}{sinpars{pi x} over
    leftlceil{x}rightrceil^{2} + leftlceil{x}rightrceil}
    ,dd x} =
    sum_{k = 0}^{infty}int_{k}^{k + 1}
    {sinpars{pi x} over
    pars{k + 1}^{2} + pars{k + 1}},dd x
    \[5mm] = &
    {2 over pi}sum_{k = 0}^{infty}
    {pars{-1}^{k} over pars{k + 1}pars{k + 2}} =
    {2 over pi}bracks{%
    sum_{k = 0}^{infty}{pars{-1}^{k} over k + 1} -
    sum_{k = 0}^{infty}{pars{-1}^{k} over k + 2}}
    \[5mm] = &
    {2 over pi}bracks{%
    -sum_{k = 1}^{infty}{pars{-1}^{k} over k} -
    sum_{k = 2}^{infty}{pars{-1}^{k} over k}}
    \[5mm] = &
    -,{2 over pi}braces{%
    sum_{k = 1}^{infty}{pars{-1}^{k} over k} +
    bracks{1 + sum_{k = 1}^{infty}{pars{-1}^{k} over k}}} \[5mm] = &
    -,{2 over pi}bracks{%
    1 + 2sum_{k = 1}^{infty}{pars{-1}^{k} over k}} =
    bbx{4lnpars{2} - 2 over pi} approx 0.2459
    end{align}






    share|cite|improve this answer



























      up vote
      8
      down vote













      $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
      newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
      newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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      newcommand{ds}[1]{displaystyle{#1}}
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      newcommand{ic}{mathrm{i}}
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      newcommand{mrm}[1]{mathrm{#1}}
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      newcommand{root}[2]{,sqrt[#1]{,{#2},},}
      newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
      newcommand{verts}[1]{leftvert,{#1},rightvert}$
      begin{align}
      &bbox[10px,#ffd]{int_{0}^{infty}{sinpars{pi x} over
      leftlceil{x}rightrceil^{2} + leftlceil{x}rightrceil}
      ,dd x} =
      sum_{k = 0}^{infty}int_{k}^{k + 1}
      {sinpars{pi x} over
      pars{k + 1}^{2} + pars{k + 1}},dd x
      \[5mm] = &
      {2 over pi}sum_{k = 0}^{infty}
      {pars{-1}^{k} over pars{k + 1}pars{k + 2}} =
      {2 over pi}bracks{%
      sum_{k = 0}^{infty}{pars{-1}^{k} over k + 1} -
      sum_{k = 0}^{infty}{pars{-1}^{k} over k + 2}}
      \[5mm] = &
      {2 over pi}bracks{%
      -sum_{k = 1}^{infty}{pars{-1}^{k} over k} -
      sum_{k = 2}^{infty}{pars{-1}^{k} over k}}
      \[5mm] = &
      -,{2 over pi}braces{%
      sum_{k = 1}^{infty}{pars{-1}^{k} over k} +
      bracks{1 + sum_{k = 1}^{infty}{pars{-1}^{k} over k}}} \[5mm] = &
      -,{2 over pi}bracks{%
      1 + 2sum_{k = 1}^{infty}{pars{-1}^{k} over k}} =
      bbx{4lnpars{2} - 2 over pi} approx 0.2459
      end{align}






      share|cite|improve this answer

























        up vote
        8
        down vote










        up vote
        8
        down vote









        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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        newcommand{ds}[1]{displaystyle{#1}}
        newcommand{expo}[1]{,mathrm{e}^{#1},}
        newcommand{ic}{mathrm{i}}
        newcommand{mc}[1]{mathcal{#1}}
        newcommand{mrm}[1]{mathrm{#1}}
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        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$
        begin{align}
        &bbox[10px,#ffd]{int_{0}^{infty}{sinpars{pi x} over
        leftlceil{x}rightrceil^{2} + leftlceil{x}rightrceil}
        ,dd x} =
        sum_{k = 0}^{infty}int_{k}^{k + 1}
        {sinpars{pi x} over
        pars{k + 1}^{2} + pars{k + 1}},dd x
        \[5mm] = &
        {2 over pi}sum_{k = 0}^{infty}
        {pars{-1}^{k} over pars{k + 1}pars{k + 2}} =
        {2 over pi}bracks{%
        sum_{k = 0}^{infty}{pars{-1}^{k} over k + 1} -
        sum_{k = 0}^{infty}{pars{-1}^{k} over k + 2}}
        \[5mm] = &
        {2 over pi}bracks{%
        -sum_{k = 1}^{infty}{pars{-1}^{k} over k} -
        sum_{k = 2}^{infty}{pars{-1}^{k} over k}}
        \[5mm] = &
        -,{2 over pi}braces{%
        sum_{k = 1}^{infty}{pars{-1}^{k} over k} +
        bracks{1 + sum_{k = 1}^{infty}{pars{-1}^{k} over k}}} \[5mm] = &
        -,{2 over pi}bracks{%
        1 + 2sum_{k = 1}^{infty}{pars{-1}^{k} over k}} =
        bbx{4lnpars{2} - 2 over pi} approx 0.2459
        end{align}






        share|cite|improve this answer














        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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        newcommand{ds}[1]{displaystyle{#1}}
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        newcommand{ic}{mathrm{i}}
        newcommand{mc}[1]{mathcal{#1}}
        newcommand{mrm}[1]{mathrm{#1}}
        newcommand{pars}[1]{left(,{#1},right)}
        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$
        begin{align}
        &bbox[10px,#ffd]{int_{0}^{infty}{sinpars{pi x} over
        leftlceil{x}rightrceil^{2} + leftlceil{x}rightrceil}
        ,dd x} =
        sum_{k = 0}^{infty}int_{k}^{k + 1}
        {sinpars{pi x} over
        pars{k + 1}^{2} + pars{k + 1}},dd x
        \[5mm] = &
        {2 over pi}sum_{k = 0}^{infty}
        {pars{-1}^{k} over pars{k + 1}pars{k + 2}} =
        {2 over pi}bracks{%
        sum_{k = 0}^{infty}{pars{-1}^{k} over k + 1} -
        sum_{k = 0}^{infty}{pars{-1}^{k} over k + 2}}
        \[5mm] = &
        {2 over pi}bracks{%
        -sum_{k = 1}^{infty}{pars{-1}^{k} over k} -
        sum_{k = 2}^{infty}{pars{-1}^{k} over k}}
        \[5mm] = &
        -,{2 over pi}braces{%
        sum_{k = 1}^{infty}{pars{-1}^{k} over k} +
        bracks{1 + sum_{k = 1}^{infty}{pars{-1}^{k} over k}}} \[5mm] = &
        -,{2 over pi}bracks{%
        1 + 2sum_{k = 1}^{infty}{pars{-1}^{k} over k}} =
        bbx{4lnpars{2} - 2 over pi} approx 0.2459
        end{align}







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 5 at 2:49

























        answered Dec 5 at 2:41









        Felix Marin

        66.8k7107139




        66.8k7107139






















            up vote
            3
            down vote













            Hint: break the integral up into
            $$int_0^infty=int_0^1+int_1^2+int_2^3+cdots.$$
            On each of these intervals the denominator takes a constant value, so you can bring it out of the integral sign and evaluate each. After that just sum all the terms you get.






            share|cite|improve this answer

















            • 1




              Can the downvoter kindly explain?
              – YiFan
              Dec 5 at 2:07















            up vote
            3
            down vote













            Hint: break the integral up into
            $$int_0^infty=int_0^1+int_1^2+int_2^3+cdots.$$
            On each of these intervals the denominator takes a constant value, so you can bring it out of the integral sign and evaluate each. After that just sum all the terms you get.






            share|cite|improve this answer

















            • 1




              Can the downvoter kindly explain?
              – YiFan
              Dec 5 at 2:07













            up vote
            3
            down vote










            up vote
            3
            down vote









            Hint: break the integral up into
            $$int_0^infty=int_0^1+int_1^2+int_2^3+cdots.$$
            On each of these intervals the denominator takes a constant value, so you can bring it out of the integral sign and evaluate each. After that just sum all the terms you get.






            share|cite|improve this answer












            Hint: break the integral up into
            $$int_0^infty=int_0^1+int_1^2+int_2^3+cdots.$$
            On each of these intervals the denominator takes a constant value, so you can bring it out of the integral sign and evaluate each. After that just sum all the terms you get.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 4 at 22:58









            YiFan

            2,1761421




            2,1761421








            • 1




              Can the downvoter kindly explain?
              – YiFan
              Dec 5 at 2:07














            • 1




              Can the downvoter kindly explain?
              – YiFan
              Dec 5 at 2:07








            1




            1




            Can the downvoter kindly explain?
            – YiFan
            Dec 5 at 2:07




            Can the downvoter kindly explain?
            – YiFan
            Dec 5 at 2:07


















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            Cabo Verde

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              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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            Gyllenstierna

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