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Integral with $(-1)^t$


Double integral with integration by partsEvaluate the integral $intfracx^2 + 1x^3 + 3x + 1 dx$How it comes that integral of odd function is not even?definite integral of a complex functionFredholm integral?finding the integral with substitutionHelp on solving integral equationCalculate the definite integral with $lim$Exstimation of a Lebesgue integralIntegral of KNN distribution













1












$begingroup$


can you suggest some exercise online where integral have $(-1)^x$ function like



$int g(x) dx$ where $g(x) = (-1)^x h(x)$ or $g(x) = h((-1)^x,x)$










share|cite|improve this question









$endgroup$











  • $begingroup$
    What is $(-1)^x$ ? It has no sense if $x$ is not an integer.
    $endgroup$
    – TheSilverDoe
    Mar 11 at 10:18






  • 3




    $begingroup$
    Or $(-1)^x$ has meaning, but complex values. Then you would probably write $exp(i pi x)$ to be clearer.
    $endgroup$
    – GEdgar
    Mar 11 at 10:28















1












$begingroup$


can you suggest some exercise online where integral have $(-1)^x$ function like



$int g(x) dx$ where $g(x) = (-1)^x h(x)$ or $g(x) = h((-1)^x,x)$










share|cite|improve this question









$endgroup$











  • $begingroup$
    What is $(-1)^x$ ? It has no sense if $x$ is not an integer.
    $endgroup$
    – TheSilverDoe
    Mar 11 at 10:18






  • 3




    $begingroup$
    Or $(-1)^x$ has meaning, but complex values. Then you would probably write $exp(i pi x)$ to be clearer.
    $endgroup$
    – GEdgar
    Mar 11 at 10:28













1












1








1





$begingroup$


can you suggest some exercise online where integral have $(-1)^x$ function like



$int g(x) dx$ where $g(x) = (-1)^x h(x)$ or $g(x) = h((-1)^x,x)$










share|cite|improve this question









$endgroup$




can you suggest some exercise online where integral have $(-1)^x$ function like



$int g(x) dx$ where $g(x) = (-1)^x h(x)$ or $g(x) = h((-1)^x,x)$







real-analysis integration complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 11 at 10:10









user1828958user1828958

526




526











  • $begingroup$
    What is $(-1)^x$ ? It has no sense if $x$ is not an integer.
    $endgroup$
    – TheSilverDoe
    Mar 11 at 10:18






  • 3




    $begingroup$
    Or $(-1)^x$ has meaning, but complex values. Then you would probably write $exp(i pi x)$ to be clearer.
    $endgroup$
    – GEdgar
    Mar 11 at 10:28
















  • $begingroup$
    What is $(-1)^x$ ? It has no sense if $x$ is not an integer.
    $endgroup$
    – TheSilverDoe
    Mar 11 at 10:18






  • 3




    $begingroup$
    Or $(-1)^x$ has meaning, but complex values. Then you would probably write $exp(i pi x)$ to be clearer.
    $endgroup$
    – GEdgar
    Mar 11 at 10:28















$begingroup$
What is $(-1)^x$ ? It has no sense if $x$ is not an integer.
$endgroup$
– TheSilverDoe
Mar 11 at 10:18




$begingroup$
What is $(-1)^x$ ? It has no sense if $x$ is not an integer.
$endgroup$
– TheSilverDoe
Mar 11 at 10:18




3




3




$begingroup$
Or $(-1)^x$ has meaning, but complex values. Then you would probably write $exp(i pi x)$ to be clearer.
$endgroup$
– GEdgar
Mar 11 at 10:28




$begingroup$
Or $(-1)^x$ has meaning, but complex values. Then you would probably write $exp(i pi x)$ to be clearer.
$endgroup$
– GEdgar
Mar 11 at 10:28










1 Answer
1






active

oldest

votes


















0












$begingroup$

I don't really know of any special sources dealing with these sorts of integrals, but I can show you how to integrate it...



First, recall that $$e^ix=cos(x)+isin(x)$$
Thus, with $x=pi$ we have $$e^ipi=-1$$
Take $ln$ on both sides to get
$$ipi=ln(-1)$$
and since $$a^x=e^xln a$$
We have that
$$(-1)^x=e^xln(-1)=e^ipi x=cos(pi x)+isin(pi x)$$
Which is easily integrated.






share|cite|improve this answer









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    1 Answer
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    active

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    1 Answer
    1






    active

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    active

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    active

    oldest

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    0












    $begingroup$

    I don't really know of any special sources dealing with these sorts of integrals, but I can show you how to integrate it...



    First, recall that $$e^ix=cos(x)+isin(x)$$
    Thus, with $x=pi$ we have $$e^ipi=-1$$
    Take $ln$ on both sides to get
    $$ipi=ln(-1)$$
    and since $$a^x=e^xln a$$
    We have that
    $$(-1)^x=e^xln(-1)=e^ipi x=cos(pi x)+isin(pi x)$$
    Which is easily integrated.






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      I don't really know of any special sources dealing with these sorts of integrals, but I can show you how to integrate it...



      First, recall that $$e^ix=cos(x)+isin(x)$$
      Thus, with $x=pi$ we have $$e^ipi=-1$$
      Take $ln$ on both sides to get
      $$ipi=ln(-1)$$
      and since $$a^x=e^xln a$$
      We have that
      $$(-1)^x=e^xln(-1)=e^ipi x=cos(pi x)+isin(pi x)$$
      Which is easily integrated.






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        I don't really know of any special sources dealing with these sorts of integrals, but I can show you how to integrate it...



        First, recall that $$e^ix=cos(x)+isin(x)$$
        Thus, with $x=pi$ we have $$e^ipi=-1$$
        Take $ln$ on both sides to get
        $$ipi=ln(-1)$$
        and since $$a^x=e^xln a$$
        We have that
        $$(-1)^x=e^xln(-1)=e^ipi x=cos(pi x)+isin(pi x)$$
        Which is easily integrated.






        share|cite|improve this answer









        $endgroup$



        I don't really know of any special sources dealing with these sorts of integrals, but I can show you how to integrate it...



        First, recall that $$e^ix=cos(x)+isin(x)$$
        Thus, with $x=pi$ we have $$e^ipi=-1$$
        Take $ln$ on both sides to get
        $$ipi=ln(-1)$$
        and since $$a^x=e^xln a$$
        We have that
        $$(-1)^x=e^xln(-1)=e^ipi x=cos(pi x)+isin(pi x)$$
        Which is easily integrated.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 11 at 19:36









        clathratusclathratus

        4,9701338




        4,9701338



























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