Differentiation under the integral sign - what transformations to use?












5












$begingroup$


Need some help with this integral



$$I (alpha) = int_1^infty {arctan(alpha x) over x^2sqrt{x^2-1}} dx$$



Taking the first derivative with respect to $alpha$



$$I' (alpha) = int_1^infty { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }$$



What transformations to use in order to solve $I'(alpha)$?






calculus integration







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    Need some help with this integral



    $$I (alpha) = int_1^infty {arctan(alpha x) over x^2sqrt{x^2-1}} dx$$



    Taking the first derivative with respect to $alpha$



    $$I' (alpha) = int_1^infty { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }$$



    What transformations to use in order to solve $I'(alpha)$?






    calculus integration







    share|cite|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$


      Need some help with this integral



      $$I (alpha) = int_1^infty {arctan(alpha x) over x^2sqrt{x^2-1}} dx$$



      Taking the first derivative with respect to $alpha$



      $$I' (alpha) = int_1^infty { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }$$



      What transformations to use in order to solve $I'(alpha)$?






      calculus integration







      share|cite|improve this question











      $endgroup$




      Need some help with this integral



      $$I (alpha) = int_1^infty {arctan(alpha x) over x^2sqrt{x^2-1}} dx$$



      Taking the first derivative with respect to $alpha$



      $$I' (alpha) = int_1^infty { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }$$



      What transformations to use in order to solve $I'(alpha)$?






      calculus integration




      calculus integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 4 hours ago









      El Pasta

      46615




      46615










      asked 4 hours ago









      KatKat

      264




      264






















          1 Answer
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          3












          $begingroup$

          Substitute
          $$u=sqrt{x^2-1}implies du=frac{x}{sqrt{x^2-1}}dximplies dx=frac{sqrt{x^2-1}}{x}du$$
          Then
          $$int { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=int { duover x^2(1+alpha^2 x^2)}=int { duover (u^2+1)(a^2u^2+a^2+1) }$$
          Perform partial fraction decomposition
          $$int { duover (u^2+1)(a^2u^2+a^2+1) }=intfrac{du}{u^2+1}-a^2intfrac{du}{a^2u^2+a^2+1}$$
          Sure you know that
          $$intfrac{du}{u^2+1}=arctan(u)+C$$
          To solve for
          $$intfrac{du}{a^2u^2+a^2+1}$$
          Use substitution
          $$v=frac{au}{sqrt{a^2+1}}implies du=frac{a^2+1}{a}$$
          $$intfrac{du}{a^2u^2+a^2+1}=intfrac{sqrt{a^2+1}dv}{a((a^2+1)v^2+a^2+1)}=frac{1}{asqrt{a^2+1}}intfrac{dv}{v^2+1}=frac{arctan(v)}{asqrt{a^2+1}}+C$$
          Now plug in back $x$, you would get



          $$int_{1}^{infty} { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=left[arctanleft(sqrt{x^2-1}right)-frac{aarctanleft(frac{asqrt{x^2-1}}{a^2+1}right)}{sqrt{a^2+1}}right]_{1}^{infty}$$
          I think you can handle the rest of the calculation.






          share|cite|improve this answer











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            1 Answer
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            1 Answer
            1






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Substitute
            $$u=sqrt{x^2-1}implies du=frac{x}{sqrt{x^2-1}}dximplies dx=frac{sqrt{x^2-1}}{x}du$$
            Then
            $$int { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=int { duover x^2(1+alpha^2 x^2)}=int { duover (u^2+1)(a^2u^2+a^2+1) }$$
            Perform partial fraction decomposition
            $$int { duover (u^2+1)(a^2u^2+a^2+1) }=intfrac{du}{u^2+1}-a^2intfrac{du}{a^2u^2+a^2+1}$$
            Sure you know that
            $$intfrac{du}{u^2+1}=arctan(u)+C$$
            To solve for
            $$intfrac{du}{a^2u^2+a^2+1}$$
            Use substitution
            $$v=frac{au}{sqrt{a^2+1}}implies du=frac{a^2+1}{a}$$
            $$intfrac{du}{a^2u^2+a^2+1}=intfrac{sqrt{a^2+1}dv}{a((a^2+1)v^2+a^2+1)}=frac{1}{asqrt{a^2+1}}intfrac{dv}{v^2+1}=frac{arctan(v)}{asqrt{a^2+1}}+C$$
            Now plug in back $x$, you would get



            $$int_{1}^{infty} { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=left[arctanleft(sqrt{x^2-1}right)-frac{aarctanleft(frac{asqrt{x^2-1}}{a^2+1}right)}{sqrt{a^2+1}}right]_{1}^{infty}$$
            I think you can handle the rest of the calculation.






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              Substitute
              $$u=sqrt{x^2-1}implies du=frac{x}{sqrt{x^2-1}}dximplies dx=frac{sqrt{x^2-1}}{x}du$$
              Then
              $$int { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=int { duover x^2(1+alpha^2 x^2)}=int { duover (u^2+1)(a^2u^2+a^2+1) }$$
              Perform partial fraction decomposition
              $$int { duover (u^2+1)(a^2u^2+a^2+1) }=intfrac{du}{u^2+1}-a^2intfrac{du}{a^2u^2+a^2+1}$$
              Sure you know that
              $$intfrac{du}{u^2+1}=arctan(u)+C$$
              To solve for
              $$intfrac{du}{a^2u^2+a^2+1}$$
              Use substitution
              $$v=frac{au}{sqrt{a^2+1}}implies du=frac{a^2+1}{a}$$
              $$intfrac{du}{a^2u^2+a^2+1}=intfrac{sqrt{a^2+1}dv}{a((a^2+1)v^2+a^2+1)}=frac{1}{asqrt{a^2+1}}intfrac{dv}{v^2+1}=frac{arctan(v)}{asqrt{a^2+1}}+C$$
              Now plug in back $x$, you would get



              $$int_{1}^{infty} { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=left[arctanleft(sqrt{x^2-1}right)-frac{aarctanleft(frac{asqrt{x^2-1}}{a^2+1}right)}{sqrt{a^2+1}}right]_{1}^{infty}$$
              I think you can handle the rest of the calculation.






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                Substitute
                $$u=sqrt{x^2-1}implies du=frac{x}{sqrt{x^2-1}}dximplies dx=frac{sqrt{x^2-1}}{x}du$$
                Then
                $$int { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=int { duover x^2(1+alpha^2 x^2)}=int { duover (u^2+1)(a^2u^2+a^2+1) }$$
                Perform partial fraction decomposition
                $$int { duover (u^2+1)(a^2u^2+a^2+1) }=intfrac{du}{u^2+1}-a^2intfrac{du}{a^2u^2+a^2+1}$$
                Sure you know that
                $$intfrac{du}{u^2+1}=arctan(u)+C$$
                To solve for
                $$intfrac{du}{a^2u^2+a^2+1}$$
                Use substitution
                $$v=frac{au}{sqrt{a^2+1}}implies du=frac{a^2+1}{a}$$
                $$intfrac{du}{a^2u^2+a^2+1}=intfrac{sqrt{a^2+1}dv}{a((a^2+1)v^2+a^2+1)}=frac{1}{asqrt{a^2+1}}intfrac{dv}{v^2+1}=frac{arctan(v)}{asqrt{a^2+1}}+C$$
                Now plug in back $x$, you would get



                $$int_{1}^{infty} { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=left[arctanleft(sqrt{x^2-1}right)-frac{aarctanleft(frac{asqrt{x^2-1}}{a^2+1}right)}{sqrt{a^2+1}}right]_{1}^{infty}$$
                I think you can handle the rest of the calculation.






                share|cite|improve this answer











                $endgroup$



                Substitute
                $$u=sqrt{x^2-1}implies du=frac{x}{sqrt{x^2-1}}dximplies dx=frac{sqrt{x^2-1}}{x}du$$
                Then
                $$int { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=int { duover x^2(1+alpha^2 x^2)}=int { duover (u^2+1)(a^2u^2+a^2+1) }$$
                Perform partial fraction decomposition
                $$int { duover (u^2+1)(a^2u^2+a^2+1) }=intfrac{du}{u^2+1}-a^2intfrac{du}{a^2u^2+a^2+1}$$
                Sure you know that
                $$intfrac{du}{u^2+1}=arctan(u)+C$$
                To solve for
                $$intfrac{du}{a^2u^2+a^2+1}$$
                Use substitution
                $$v=frac{au}{sqrt{a^2+1}}implies du=frac{a^2+1}{a}$$
                $$intfrac{du}{a^2u^2+a^2+1}=intfrac{sqrt{a^2+1}dv}{a((a^2+1)v^2+a^2+1)}=frac{1}{asqrt{a^2+1}}intfrac{dv}{v^2+1}=frac{arctan(v)}{asqrt{a^2+1}}+C$$
                Now plug in back $x$, you would get



                $$int_{1}^{infty} { dxover (1+alpha^2 x^2) xsqrt{x^2-1} }=left[arctanleft(sqrt{x^2-1}right)-frac{aarctanleft(frac{asqrt{x^2-1}}{a^2+1}right)}{sqrt{a^2+1}}right]_{1}^{infty}$$
                I think you can handle the rest of the calculation.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 3 hours ago

























                answered 3 hours ago









                LarryLarry

                2,1862828




                2,1862828






























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