How to solve $int_0^{infty}frac{x^{alpha}log^n(x)}{1+x^2}dx$?











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Recently I posted a rather similar question in here:



How to solve $int_0^{infty}frac{log^n(x)}{1+x^2}dx$?



Now I was looking for harder integrals, for which I know how to calculate the case $n=1$. Mathematica provides analtical solutions.



$$
int_0^{infty}frac{x^{alpha}log^n(x)}{1+x^2}dx=frac{1}{4^{n+1}}Gamma(n+1)left[zetaleft(n+1,frac{1-alpha}{4}right)-zetaleft(n+1,frac{3-alpha}{4}right)+(-1)^nleft(zetaleft(n+1,frac{1+alpha}{4}right)-zetaleft(n+1,frac{3+alpha}{4}right)right)right]
$$



for $ninmathbb{N}$ and $-1<alpha<1$.



$$
int_0^{infty}frac{x^{alpha}log(x)}{(ax^2+b)^n}dx
$$



for $ninmathbb{N}$ and $0<alpha+1<2n$ and $a,b>0$.
How to deal with the integrals at hand?






integration complex-analysis







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  • 1




    This is very similar to this question and the solution here can be gotten by differentiating $n$ times instead of once, as in this answer.
    – robjohn♦
    Nov 23 at 15:08















up vote
2
down vote

favorite
2












Recently I posted a rather similar question in here:



How to solve $int_0^{infty}frac{log^n(x)}{1+x^2}dx$?



Now I was looking for harder integrals, for which I know how to calculate the case $n=1$. Mathematica provides analtical solutions.



$$
int_0^{infty}frac{x^{alpha}log^n(x)}{1+x^2}dx=frac{1}{4^{n+1}}Gamma(n+1)left[zetaleft(n+1,frac{1-alpha}{4}right)-zetaleft(n+1,frac{3-alpha}{4}right)+(-1)^nleft(zetaleft(n+1,frac{1+alpha}{4}right)-zetaleft(n+1,frac{3+alpha}{4}right)right)right]
$$



for $ninmathbb{N}$ and $-1<alpha<1$.



$$
int_0^{infty}frac{x^{alpha}log(x)}{(ax^2+b)^n}dx
$$



for $ninmathbb{N}$ and $0<alpha+1<2n$ and $a,b>0$.
How to deal with the integrals at hand?






integration complex-analysis







share|cite|improve this question


















  • 1




    This is very similar to this question and the solution here can be gotten by differentiating $n$ times instead of once, as in this answer.
    – robjohn♦
    Nov 23 at 15:08













up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2





Recently I posted a rather similar question in here:



How to solve $int_0^{infty}frac{log^n(x)}{1+x^2}dx$?



Now I was looking for harder integrals, for which I know how to calculate the case $n=1$. Mathematica provides analtical solutions.



$$
int_0^{infty}frac{x^{alpha}log^n(x)}{1+x^2}dx=frac{1}{4^{n+1}}Gamma(n+1)left[zetaleft(n+1,frac{1-alpha}{4}right)-zetaleft(n+1,frac{3-alpha}{4}right)+(-1)^nleft(zetaleft(n+1,frac{1+alpha}{4}right)-zetaleft(n+1,frac{3+alpha}{4}right)right)right]
$$



for $ninmathbb{N}$ and $-1<alpha<1$.



$$
int_0^{infty}frac{x^{alpha}log(x)}{(ax^2+b)^n}dx
$$



for $ninmathbb{N}$ and $0<alpha+1<2n$ and $a,b>0$.
How to deal with the integrals at hand?






integration complex-analysis







share|cite|improve this question













Recently I posted a rather similar question in here:



How to solve $int_0^{infty}frac{log^n(x)}{1+x^2}dx$?



Now I was looking for harder integrals, for which I know how to calculate the case $n=1$. Mathematica provides analtical solutions.



$$
int_0^{infty}frac{x^{alpha}log^n(x)}{1+x^2}dx=frac{1}{4^{n+1}}Gamma(n+1)left[zetaleft(n+1,frac{1-alpha}{4}right)-zetaleft(n+1,frac{3-alpha}{4}right)+(-1)^nleft(zetaleft(n+1,frac{1+alpha}{4}right)-zetaleft(n+1,frac{3+alpha}{4}right)right)right]
$$



for $ninmathbb{N}$ and $-1<alpha<1$.



$$
int_0^{infty}frac{x^{alpha}log(x)}{(ax^2+b)^n}dx
$$



for $ninmathbb{N}$ and $0<alpha+1<2n$ and $a,b>0$.
How to deal with the integrals at hand?






integration complex-analysis




integration complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Nov 23 at 14:29









Schnarco

1428




1428








  • 1




    This is very similar to this question and the solution here can be gotten by differentiating $n$ times instead of once, as in this answer.
    – robjohn♦
    Nov 23 at 15:08














  • 1




    This is very similar to this question and the solution here can be gotten by differentiating $n$ times instead of once, as in this answer.
    – robjohn♦
    Nov 23 at 15:08








1




1




This is very similar to this question and the solution here can be gotten by differentiating $n$ times instead of once, as in this answer.
– robjohn♦
Nov 23 at 15:08




This is very similar to this question and the solution here can be gotten by differentiating $n$ times instead of once, as in this answer.
– robjohn♦
Nov 23 at 15:08










3 Answers
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Referring to this answer and using the definition of $I(a)$ in it,
$$int^infty_0frac{x^alog^n(x)}{x^2+1}dx=I^{(n)}(a)=left(frac{pi}2right)^{n+1}sec^{(n)}left(frac{pi a}2right)$$






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    up vote
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    down vote













    By considering
    $$ frac{1}{2pi i} int frac{z^a}{1+z^2} , dz $$
    around the standard keyhole contour. Then differentiate wrt $a$, $n$ times, and set $a=alpha$.






    share|cite|improve this answer






























      up vote
      0
      down vote













      In the $n=0$ case, substituting $x=tan t$ gives $$int_0^inftytfrac{x^alpha dx}{1+x^2}=int_0^inftytan^alpha dx=tfrac{1}{2}operatorname{B}(tfrac{1+alpha}{2},,tfrac{1-alpha}{2})=tfrac{pi}{2}csc(tfrac{pi}{2}+tfrac{pialpha}{2})=tfrac{pi}{2}sec(tfrac{pialpha}{2}).$$Applying $partial_a^n$,$$int_0^inftytfrac{x^alphaln^nxdx}{1+x^2}=(tfrac{pi}{2})^{n+1}sec^{(n)}(tfrac{pialpha}{2}).$$






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote



        accepted










        Referring to this answer and using the definition of $I(a)$ in it,
        $$int^infty_0frac{x^alog^n(x)}{x^2+1}dx=I^{(n)}(a)=left(frac{pi}2right)^{n+1}sec^{(n)}left(frac{pi a}2right)$$






        share|cite|improve this answer

























          up vote
          2
          down vote



          accepted










          Referring to this answer and using the definition of $I(a)$ in it,
          $$int^infty_0frac{x^alog^n(x)}{x^2+1}dx=I^{(n)}(a)=left(frac{pi}2right)^{n+1}sec^{(n)}left(frac{pi a}2right)$$






          share|cite|improve this answer























            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Referring to this answer and using the definition of $I(a)$ in it,
            $$int^infty_0frac{x^alog^n(x)}{x^2+1}dx=I^{(n)}(a)=left(frac{pi}2right)^{n+1}sec^{(n)}left(frac{pi a}2right)$$






            share|cite|improve this answer












            Referring to this answer and using the definition of $I(a)$ in it,
            $$int^infty_0frac{x^alog^n(x)}{x^2+1}dx=I^{(n)}(a)=left(frac{pi}2right)^{n+1}sec^{(n)}left(frac{pi a}2right)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 23 at 14:47









            Szeto

            6,3092826




            6,3092826






















                up vote
                1
                down vote













                By considering
                $$ frac{1}{2pi i} int frac{z^a}{1+z^2} , dz $$
                around the standard keyhole contour. Then differentiate wrt $a$, $n$ times, and set $a=alpha$.






                share|cite|improve this answer



























                  up vote
                  1
                  down vote













                  By considering
                  $$ frac{1}{2pi i} int frac{z^a}{1+z^2} , dz $$
                  around the standard keyhole contour. Then differentiate wrt $a$, $n$ times, and set $a=alpha$.






                  share|cite|improve this answer

























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    By considering
                    $$ frac{1}{2pi i} int frac{z^a}{1+z^2} , dz $$
                    around the standard keyhole contour. Then differentiate wrt $a$, $n$ times, and set $a=alpha$.






                    share|cite|improve this answer














                    By considering
                    $$ frac{1}{2pi i} int frac{z^a}{1+z^2} , dz $$
                    around the standard keyhole contour. Then differentiate wrt $a$, $n$ times, and set $a=alpha$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 23 at 14:52

























                    answered Nov 23 at 14:38









                    Richard Martin

                    1,63918




                    1,63918






















                        up vote
                        0
                        down vote













                        In the $n=0$ case, substituting $x=tan t$ gives $$int_0^inftytfrac{x^alpha dx}{1+x^2}=int_0^inftytan^alpha dx=tfrac{1}{2}operatorname{B}(tfrac{1+alpha}{2},,tfrac{1-alpha}{2})=tfrac{pi}{2}csc(tfrac{pi}{2}+tfrac{pialpha}{2})=tfrac{pi}{2}sec(tfrac{pialpha}{2}).$$Applying $partial_a^n$,$$int_0^inftytfrac{x^alphaln^nxdx}{1+x^2}=(tfrac{pi}{2})^{n+1}sec^{(n)}(tfrac{pialpha}{2}).$$






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          In the $n=0$ case, substituting $x=tan t$ gives $$int_0^inftytfrac{x^alpha dx}{1+x^2}=int_0^inftytan^alpha dx=tfrac{1}{2}operatorname{B}(tfrac{1+alpha}{2},,tfrac{1-alpha}{2})=tfrac{pi}{2}csc(tfrac{pi}{2}+tfrac{pialpha}{2})=tfrac{pi}{2}sec(tfrac{pialpha}{2}).$$Applying $partial_a^n$,$$int_0^inftytfrac{x^alphaln^nxdx}{1+x^2}=(tfrac{pi}{2})^{n+1}sec^{(n)}(tfrac{pialpha}{2}).$$






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            In the $n=0$ case, substituting $x=tan t$ gives $$int_0^inftytfrac{x^alpha dx}{1+x^2}=int_0^inftytan^alpha dx=tfrac{1}{2}operatorname{B}(tfrac{1+alpha}{2},,tfrac{1-alpha}{2})=tfrac{pi}{2}csc(tfrac{pi}{2}+tfrac{pialpha}{2})=tfrac{pi}{2}sec(tfrac{pialpha}{2}).$$Applying $partial_a^n$,$$int_0^inftytfrac{x^alphaln^nxdx}{1+x^2}=(tfrac{pi}{2})^{n+1}sec^{(n)}(tfrac{pialpha}{2}).$$






                            share|cite|improve this answer












                            In the $n=0$ case, substituting $x=tan t$ gives $$int_0^inftytfrac{x^alpha dx}{1+x^2}=int_0^inftytan^alpha dx=tfrac{1}{2}operatorname{B}(tfrac{1+alpha}{2},,tfrac{1-alpha}{2})=tfrac{pi}{2}csc(tfrac{pi}{2}+tfrac{pialpha}{2})=tfrac{pi}{2}sec(tfrac{pialpha}{2}).$$Applying $partial_a^n$,$$int_0^inftytfrac{x^alphaln^nxdx}{1+x^2}=(tfrac{pi}{2})^{n+1}sec^{(n)}(tfrac{pialpha}{2}).$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 23 at 14:58









                            J.G.

                            21.1k21933




                            21.1k21933






























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