Vtyjkyuk

If $displaystyleint_0^inftydfracx^3x^2+a^2,mathrmdx=largedisplaystyledfrac1ka^6$, then find the value of $displaystyledfrack8$.

I tried a lot but finally stuck at an intermediate form :

$$beginalign
&int_0^inftydfracx^3x^2+a^2,mathrmdx, textwith, x^2=t,2x~mathrmdx=mathrmdt\
&=frac12int_0^inftydfrac(x^2)(2x)x^2+a^2,mathrmdx=frac12int_0^inftydfractt+a^2,mathrmdt=frac12int_0^inftydfract+a^2-a^2t+a^2,mathrmdt\
&=frac12left[int_0^inftymathrmdt-int_0^inftydfraca^2t+a^2,mathrmdtright]=frac12left[t|_0^infty-a^2ln(a^2+t)|_0^inftyright]
endalign$$

$$I=displaystyleintdfracx^3x^2+a^2,mathrmdx$$

Substitute $u=x^2+a^2$ thus $mathrmdx=dfrac12x,mathrmdu$

$$I=classsteps-nodecssIdsteps-node-1dfrac12displaystyleintdfracu-a^2u,mathrmdu$$

$$I=dfrac12displaystyleintleft(1-dfraca^2uright)mathrmdu$$
$$I=dfrac12displaystyleint1,mathrmdu-dfrac12classsteps-nodecssIdsteps-node-2a^2displaystyleintdfrac1u,mathrmdu$$
$$I=dfracu2-dfraca^2lnleft(uright)2+c$$
$$I=left(dfracx^2+a^22-dfraca^2lnleft(x^2+a^2right)2right)biggr|_x=0^infty$$

The integral is divergent.

$$I=int_0^inftyfracx^3x^2+a^2dx$$
firstly, let $u=x^2+a^2$ so $fracdudx=2x therefore dx=fracdu2x$

For $x=0$, $u=a^2$. For $xtoinfty$, $lim_xto inftyleft(x^2+a^2right)toinfty$
so our integral becomes:

$$I=frac12int_a^2^inftyfracx^3x^2+a^2.fracdux=frac12int_a^2^inftyfracx^2x^2+a^2dx=frac12int_a^2^inftyfracu-a^2udu=frac12int_a^2^infty du-fraca^22int_a^2^inftyfrac1udu$$

Both parts to this integral are divergent and so the integral cannot be calculated

The integral does not converge.

Let

$$ I=int_0^inftydfracx^3x^2+a^2,mathrmdx.$$

By the Schwinger parametrization we have

$$ I=int_0^infty mathrmdt, exp(-ta^2)int_0^infty mathrmdx, x^3 expleft(-tx^2right).$$

The last integral can be calculated by the Feynman trick. Using this result, one gets

$$I=int_0^infty mathrmdt,frac sqrtpiexp(-ta^2)4 t^3/2=fracsqrtpia4Gamma left(-1,0right),$$

which is infinity since

$$ Gammaleft(-1,0right)equivint_0^infty mathrmdt, t^-3/2exp(-t)=tildeinfty.$$