While reading a number theory paper I encountered the identity

$$ \int_{- \infty}^{\infty} (1 + x^2)^{ - \frac{z}{2} - 1} dx = \sqrt{\pi} \frac{ \Gamma(\frac{z + 1}{2}) }{\Gamma(\frac{z}{2} + 1)},$$

apparently true for all $z \in \bf{C}$ for which the integral on the left converges absolutely. The author offers neither a justification nor a citation, so apparently this is 'well known'.

I thought of two ideas for how to prove it: (1) Use integration by parts to prove that both sides satisfy the same functional equation, and then attempt to adapt the proof of the Bohr-Mollerup theorem to prove that this functional equation characterizes the function; (2) use integration by parts to analytically continue the left-side function in $z$, determine all the poles and their residues, and attempt to use Weierstrass factorization to prove that this characterizes the function.

Both approaches looked rather involved and I didn't attempt to work out the details of either. Is this, indeed, 'well known'? Is there a suitable reference which proves this, along with other identities of this nature? And to what extent does the identity generalize?

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    $\begingroup$ Did you look in Gradshteyn and Ryzhik? $\endgroup$
    – Lucia
    Commented Jun 10, 2016 at 19:29
  • $\begingroup$ @Lucia: I didn't; I will take your comment as advice that I should! $\endgroup$ Commented Jun 10, 2016 at 19:44
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    $\begingroup$ Presumably one starts with the identity $\int_0^\infty e^{-at} t^s \frac{dt}{t} = \Gamma(s) a^{-s}$ with $a = 1+x^2$ and $s = \frac{z}{2}+1$, and then integrates in $x$ using Fubini's theorem? $\endgroup$
    – Terry Tao
    Commented Jun 10, 2016 at 19:46
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    $\begingroup$ Basically, Gamma functions show up whenever one tries to write multiplicative characters such as $a \mapsto a^{-s}$ in terms of additive characters such as $a \mapsto e^{-at}$. Here, $x \mapsto e^{-t(1+x^2)}$ is significantly easier to integrate than $x \mapsto (1+x^2)^{-\frac{z}{2}-1}$, so that suggests the use of the Gamma function to transform the latter into a suitable average of the former. The same trick can be used to prove the functional equation: terrytao.wordpress.com/2014/12/15/… $\endgroup$
    – Terry Tao
    Commented Jun 10, 2016 at 19:53
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    $\begingroup$ Regarding a reference, the particular identity in question can be deduced from the general properties of Bessel potentials: en.wikipedia.org/wiki/Bessel_potential . Stein's "Singular integrals..." is a good reference. $\endgroup$
    – Terry Tao
    Commented Jun 10, 2016 at 20:40

2 Answers 2


Yes, there is a trick which generalizes to analogous integrals on the classical cones, using the Gamma functions attached to these cones. In this, the simplest case, the starting point is the observation that $\int_0^\infty e^{-ty}\,t^s\;{dt\over t}=y^{-s}\cdot \Gamma(s)$ for $y>0$, and then for $\Re(y)>0$ by analytic continuation. This identity has the obvious analogues for other classical cones, such as positive-definite symmetric real $n$-by-$n$ matrices, or light cones, etc.

Thus, your given integral is ${1\over \Gamma(z/2+1)} \int_0^\infty\int_{\mathbb R} t^{z/2+1}\,e^{-t(x^2+1)}\;dx\;{dt\over t}$. Integrating the Gaussian in $x$ first gives ${1\over \Gamma(z/2+1)} \int_0^\infty t^{{z+1\over 2}}\,e^{-t}\;dx\;{dt\over t}$.

EDIT: While we're here, it's probably reasonable to give some further explicit details, as well as historical context. First, historically, C. L. Siegel's 1939 paper on "Siegel modular forms" did broach issues similar to the present. Among others along the way, G. Shimura's papers in the early 1980's used such integral identities quite aggressively, and proved (among other things) identities generalizing the classical ones involving confluent hypergeometric (a.k.a. Bessel) functions.

An example of a less trivial identity: let $C_n$ be the cone of positive-definite symmetric real $n$-by-$n$ matrices, and $\Gamma_C(s)=\int_{C_n} (\det t)^s\,e^{-{\mathrm tr}\, t}\,dt/\det^nt$. On one hand, by an induction, we can show that this is a product of ordinary Gammas $\Gamma(s)\Gamma(s-1/2)\Gamma(s-1)\Gamma(s-3/2)... \Gamma(s-(n-1)/2)$, by fooling around with changes of variables. On the other hand, in parallel to the much-better-known identity for the cone $(0,\infty)$, we have $\int_{C_n} (\det t)^s\,e^{{\mathrm tr}\,ty}\,dt/\det t^n=(\det y)^{-s}\Gamma_C(s)$, by the slighly subtler change of variables replacing $t$ by $y^{-1/2}ty^{-1/2}$, for $y\in C_n$. And so on...


It is nothing but beta-function. Consider only positive $x$ and denote $1/(x^2+1)=t$. You get $$\int_0^\infty (1+x^2)^{-z/2-1}dx=\frac12 \int_0^1 t^{z/2-1/2}(1-t)^{-1/2}dt=\frac12 B((z+1)/2,1/2)=\\=\frac{\Gamma(1/2)\Gamma((z+1)/2)}{2\Gamma(z/2+1)}=\sqrt{\pi}\frac{\Gamma((z+1)/2)}{2\Gamma(z/2+1)}.$$

I think, there should be a reference for all integrals of the form $\int_0^\infty x^a (1+x^b)^{-c} dx$, which reduce to the beta-function after the change of variables $1/(1+x^b)=t$. At least it is surely well-known. In Demidovich problems book it is ex.3853, for example.

  • $\begingroup$ I checked that this works too; thank you very much! $\endgroup$ Commented Jun 10, 2016 at 21:45
  • $\begingroup$ It is also possible to apply the Slater theorem - a general key to such integrals. $\endgroup$
    – Sergei
    Commented Jun 14, 2016 at 13:45

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