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If $\displaystyle\int_0^{\infty}\dfrac{x^3}{x^2+a^2}\,\mathrm{d}x=\large\displaystyle\dfrac1{ka^6}$, then find the value of $\displaystyle\dfrac{k}{8}$.

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

$$\begin{align} &\int_0^{\infty}\dfrac{x^3}{x^2+a^2}\,\mathrm{d}x, \text{with}\, {x^2=t},{2x~\mathrm{d}x=\mathrm{d}t}\\ &=\frac12\int_0^{\infty}\dfrac{(x^2)(2x)}{x^2+a^2}\,\mathrm{d}x=\frac12\int_0^{\infty}\dfrac{t}{t+a^2}\,\mathrm{d}t=\frac12\int_0^{\infty}\dfrac{t+a^2-a^2}{t+a^2}\,\mathrm{d}t\\ &=\frac12\left[\int_0^{\infty}\mathrm{d}t-\int_0^{\infty}\dfrac{a^2}{t+a^2}\,\mathrm{d}t\right]=\frac12\left[t|_0^{\infty}-a^2\ln(a^2+t)|_0^{\infty}\right] \end{align}$$

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    $\begingroup$ Please refrain from posting photos and make a post using MathJax so that's easily readable but also searchable. $\endgroup$ – Rebellos Sep 9 '18 at 8:46
  • $\begingroup$ Your derivation seems to be correct and the integral diverges, just collect t in the last expression and then take the limit $\to \infty$. $\endgroup$ – gimusi Sep 9 '18 at 8:49
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    $\begingroup$ Your anti-derivative is correct. The integral diverges - this can be seen without computation, because the function is continuous on $[0,\infty$ and behaves asymptotically like $x.$ $\endgroup$ – user376343 Sep 9 '18 at 8:56
  • $\begingroup$ $$I=\int_0^\infty\frac{x^3}{x^2+a^2}dx$$ firstly, let $u=x^2+a^2$ so $\frac{du}{dx}=2x \therefore dx=\frac{du}{2x}$ For $x=0$, $u=a^2$. For $x\to\infty$, $\lim_{x\to \infty}\left(x^2+a^2\right)\to\infty$ so our integral becomes: $$I=\frac{1}{2}\int_{a^2}^\infty\frac{x^3}{x^2+a^2}.\frac{du}{x}=\frac{1}{2}\int_{a^2}^\infty\frac{x^2}{x^2+a^2}dx=\frac{1}{2}\int_{a^2}^\infty\frac{u-a^2}{u}du=\frac{1}{2}\int_{a^2}^\infty du-\frac{a^2}{2}\int_{a^2}^\infty\frac{1}{u}du$$ Both parts to this integral are divergent and so the integral cannot be calculated $\endgroup$ – Henry Lee Sep 9 '18 at 15:53
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$$I={\displaystyle\int}\dfrac{x^3}{x^2+a^2}\,\mathrm{d}x$$

Substitute $u=x^2+a^2$ thus $\mathrm{d}x=\dfrac{1}{2x}\,\mathrm{d}u$

$$I=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{u-a^2}{u}\,\mathrm{d}u$$

$$I={\dfrac{1}{2}\displaystyle\int}\left(1-\dfrac{a^2}{u}\right)\mathrm{d}u$$ $$I=\dfrac{1}{2}{\displaystyle\int}1\,\mathrm{d}u-\dfrac{1}{2}\class{steps-node}{\cssId{steps-node-2}{a^2}}{\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u$$ $$I=\dfrac{u}{2}-\dfrac{a^2\ln\left(u\right)}{2}+c$$ $$I=\left(\dfrac{x^2+a^2}{2}-\dfrac{a^2\ln\left(x^2+a^2\right)}{2}\right)\biggr|_{x=0}^{\infty}$$

The integral is divergent.

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The integral does not converge.

Let

$$ I=\int_0^{\infty}\dfrac{x^3}{x^2+a^2}\,\mathrm{d}x.$$

By the Schwinger parametrization we have

$$ I=\int_0^\infty \mathrm{d}t\, \exp(-ta^2)\int_0^\infty \mathrm{d}x\, x^3 \exp\left(-tx^2\right).$$

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

$$I=\int_0^\infty \mathrm{d}t\,\frac{ \sqrt{\pi}\exp(-ta^2)}{4 t^{3/2}}=\frac{\sqrt{\pi}a}{4}\Gamma \left(-1,0\right),$$

which is infinity since

$$ \Gamma\left(-1,0\right)\equiv\int_0^\infty \mathrm{d}t\, t^{-3/2}\exp(-t)=\tilde{\infty}.$$

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$$I=\int_0^\infty\frac{x^3}{x^2+a^2}dx$$ firstly, let $u=x^2+a^2$ so $\frac{du}{dx}=2x \therefore dx=\frac{du}{2x}$

For $x=0$, $u=a^2$. For $x\to\infty$, $\lim_{x\to \infty}\left(x^2+a^2\right)\to\infty$ so our integral becomes:

$$I=\frac{1}{2}\int_{a^2}^\infty\frac{x^3}{x^2+a^2}.\frac{du}{x}=\frac{1}{2}\int_{a^2}^\infty\frac{x^2}{x^2+a^2}dx=\frac{1}{2}\int_{a^2}^\infty\frac{u-a^2}{u}du=\frac{1}{2}\int_{a^2}^\infty du-\frac{a^2}{2}\int_{a^2}^\infty\frac{1}{u}du$$

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

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