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This is another integral in the book "irresistible integral" I can find that: $$\int_0^\infty \frac{1}{\left( x^4 +2ax^2+1 \right)^{m+1}} \, \text{d}x =\frac{{{\left( -1 \right)}^m}\sqrt{2}\pi }{4\left( m! \right)}\left. \left( \frac{\text{d}^m}{\text{d}x^m} \sqrt{x}\sqrt{a+\sqrt{x}} \right) \right|_{x=1}$$ Do we have a better closed form? or better solution? My working: Let $$I(b)=\int_0^\infty \frac{1}{x^4+2ax^2+b}\,\text{d}x$$ then $$I^{(m)}(b)=m!(-1)^m\int_0^\infty \frac{1}{(x^4+2ax^2+b)^{m+1}} \, \text{d}x$$ Also $$I(b)=\int_0^\infty \frac{1}{(x^2+\alpha)(x^2+\beta)}\,\text{d}x$$ where $$\alpha=a+\sqrt{a^2-b}>0,\ \ \beta=a-\sqrt{a^2-b}>0$$ $$I(b)=\frac{1}{\beta-\alpha}\int_0^\infty \left(\frac{1}{x^2+\alpha}-\frac{1}{x^2+\beta}\right)$$ $$=\frac{\pi}{2(\beta-\alpha)}\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\sqrt{\beta}}\right)$$ $$=\frac{\pi}{4\sqrt{a^2-b}}\frac{\sqrt{2}\sqrt{a-\sqrt{b}}}{\sqrt{b}} = \frac{\sqrt{2}\pi}{4\sqrt{b}\sqrt{a+\sqrt{b}}}$$ Which follows that $$\int_0^\infty \frac{1}{( x^4+2ax^2+1 )^{m+1}}\text{d}x =\frac{(-1)^m\sqrt{2}\pi }{4m!} \left. \left( \frac{\text{d}^m}{\text{d}x^m}\sqrt{x}\sqrt{a+\sqrt{x}} \right) \right|_{x=1}$$

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As far as solutions go, this seems perfectly reasonable. It uses nothing more than partial fractions and differentiation under the integral. For the answer, though, I agree that it's a little cumbersome to be called closed form, since it requires multiple differentiations to actually evaluate. Have you tried doing the evaluation? How horrible does the evaluated answer look? It is possible that what you have is significantly better than what you will get. – Aaron Feb 11 '13 at 13:50
I changed things like \int{\frac{{{1}}}{{\left({{x}^{4}}+2{{a}{x}^{{2}}+{1}\right)^{m+1}}}\mathrm{d}{x‌​}} to \int\frac{1}{(x^4+2ax^2+1)^{m+1}\,\mathrm{d}x. It approached a parody of atrocious taste. Anyone trying to see how to use TeX by looking at this will think it's hideously complicated. I realize one blames whatever weird software is used by people who post stuff like this, but maybe we should complain to whoever writes that software. – Michael Hardy Feb 11 '13 at 14:21
up vote 2 down vote accepted

For every $a\gt-1$, consider $$ I_m(a)=\int_0^\infty \frac{\mathrm dx}{\left(x^4+2ax^2+1\right)^{m+1}},$$ then, for every $|t|$ small enough, $$ \sum_{m=0}^{+\infty}I_m(a)t^m=\frac{\pi}{2\sqrt2\sqrt{1-t}\sqrt{a+\sqrt{1-t}}}. $$ To deduce a closed form formula for each $I_m(a)$ from this expression does not seem obvious. However, one sees readily that the radius of convergence of the series in the LHS is $1-a^2$ if $-1\lt a\leqslant0$ and $1$ if $a\geqslant0$. Thus, when $m\to\infty$, $m^{-1}\log I_m(a)\to-\log(1-a^2)$ if $-1\lt a\leqslant0$ and $m^{-1}\log I_m(a)\to0$ if $a\geqslant0$.

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@MichaelHardy You modified my post to replace \int_{0}^{\infty} by \int_0^\infty. No doubt you have tons of exceedingly good reasons to prefer one over the other. Nevertheless I wish that in the future you exercise your typographical talents elsewhere than in my answers, as long as these concern such petty matters. Thanks in advance. – Did Feb 11 '13 at 14:41
Did (86.8k) vs. Michael Hardy (41.4k): clash of the Titans. – Matemáticos Chibchas Feb 11 '13 at 15:15

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