$\int^{\infty}_{0}x^{r +s- 1}(1 + x)^{-s}(1 + x^2)^{-\frac{rm}{2}}dx$

Question

I'm trying to solve the integral

$\int^{\infty}_{0}x^{r +s- 1}(1 + x)^{-s}(1 + x^2)^{-\frac{rm}{2}}dx$,

where $s$, $r$ and $m$>1 are positive integers.

My question is whether a closed form solution for this integral exists. By closed form here I mean an expression in terms of a finite number of special functions. There is some hope for a closed form solution I think given the output below from symbolic software, but I have been unable to find a formula yet.

To give a bit of context the integral is the $s$-th moment of the real random variable $x/(1+x)$, $x>0$, when the probability density function of $x$ is, up to a constant that I have omitted for simplicity, is $x^{r - 1}(1 + x^2)^{-\frac{rm}{2}}$. All these moments exist. I am trying to understand whether they are expressible in terms of a finite number of special functions.

I was not able to get help from Mathematica or Maple. Both Mathematica and Maple give me a solution containing a $\Gamma$ function evaluated at negative integers. More precisely, if I input in Mathematica:

Integrate[ x^(r + s - 1) (1 + x)^-s (1 + x^2)^(-r m/2), {x, 0, \[Infinity]}, Assumptions -> {r \[Element] Integers, s \[Element] Integers, m \[Element] Integers, r > 0, s > 0, m > 1}]

I obtain:

(1/Gamma[s]) Gamma[(-1 + m) r] Gamma[ r - m r + s] HypergeometricPFQ[{(m r)/2, -(r/2) + (m r)/2, 1/2 - r/2 + (m r)/2}, {1/2 - r/2 + (m r)/2 - s/2, 1 - r/2 + (m r)/2 - s/2}, -1] + (1/( 2 Gamma[(m r)/ 2]))(-s Gamma[1/2 (-1 + (-1 + m) r - s)] Gamma[ 1/2 (1 + r + s)] HypergeometricPFQ[{1/2 + s/2, 1 + s/2, 1/2 + r/2 + s/2}, {3/2, 3/2 + r/2 - (m r)/2 + s/2}, -1] + Gamma[1/2 ((-1 + m) r - s)] Gamma[(r + s)/ 2] HypergeometricPFQ[{1/2 + s/2, r/2 + s/2, s/2}, {1/2, 1 + r/2 - (m r)/2 + s/2}, -1])

which is:

$\frac{\Gamma(r(m-1))\Gamma(s-r(m-1))}{\Gamma(s)}{}_3 F_2\left( \frac {mr}{2},\frac{r(m-1)}{2},\frac{1+r(m-1)}{2};\frac{1-s+r(m-1)}{2},1+\frac {s-r(m+1)}{2};-1\right)$ $ -\frac{s\Gamma\left(\frac{r(m-1)-1-s}{2}\right)\Gamma\left( \frac{1+r+s}{2}\right) } {2\Gamma(\frac{mr}{2})}{}_3 F_2\left( \frac{1+s}{2},1+\frac{s}{2} ,\frac{1+r+s}{2};\frac{3}{2},\frac{3+s-r(m-1)}{2};-1\right)$ $ +\Gamma\left( \frac{r(m-1)-s}{2}\right) \Gamma\left( \frac{r+s} {2}\right){}_3 F_2\left( \frac{1+s}{2},\frac{r+s}{2},\frac{s}{2} ;\frac{1}{2},1+\frac{s-r(m-1)}{2};-1\right)$

(same solution for Maple) As you can see, there is a $\Gamma$ function evaluated at $s-r(m-1)$ which can be a negative integer, so it seems to me that Mathematica and Maple are giving a solution that holds for most real values of $(r,m,s)$ but not necessarily for the values I'm interested in ($r,m,s$ are positive integers in my problem)

I'm not too sure my question is suitable here, but I did not have any luck at math.stackexchange (https://math.stackexchange.com/questions/234989/int-infty-0xr-s-11-x-s1-x2-fracrm2dx) so I thought I'd try and see if I can get some help here.

Answer 1

Suppose $r$ is even, and write $r = 2k$ and $rm/2 = k+j$. Let $$ J(s,j,k) = \int_0^\infty \frac{x^{2k+s-1}}{(1+x)^s(1+x^2)^{k+j}}\ dx$$ The integral converges if $j \ge 1$ and $s \ge 1$. The (ordinary) generating function of this is $$G(u,v,w) = \sum_{j=1}^\infty \sum_{k=0}^\infty \sum_{s=1}^\infty J(s,j,k) u^j v^k w^s$$ Now for $|u|,|v|,|w|<1$ $$ \sum_{j=1}^\infty \sum_{k=0}^\infty \sum_{s=1}^\infty \frac{x^{2k+s-1} u^jv^k w^s}{(1+x)^s(1+x^2)^{k+j}} = \frac{u w (1+x^2)}{ (1-u+x^2)(1 + (1-v)x^2)(1+(1-w)x)}$$ Integrating on $(0,\infty)$, Maple finds (for $|u|,|v|,|w|<1$) a rather formidable closed-form expression for $G(u,v,w)$ which is a rational function in $u$, $v$, $w$, $\sqrt{1-u}$, $\sqrt{1-v}$, $\ln(1-u)$, $\ln(1-v)$ and $\ln(1-w)$.

Answer 2

Rather than compute that integral directly, perturb it a little. In this case, multiply it by $x^{\alpha}$ with $\alpha > 0$. Maple can compute the answer (I guess Mathematica can too). I do not post the answer here because it is too much wallpaper.

Then you can examine what happens when $\alpha \rightarrow 0$.

The appearance of a lot more terms in the perturbed version gives an idea of all the things that collapse at $\alpha=0$. The hope is then that, for generic $\alpha$ close to $0$, the result does not in fact involve any $\Gamma$ evaluated at negative integer arguments. Then you can figure out what really happens in these cases, by first specializing your parameters $r,s,m$ so that the original would be problematic, then take the limit.

Of course you can also try $e^{-\alpha x}$ as a perturbation, but (in Maple), I had no luck with that. Maybe Mathematica can handle that one.

Answer 3

There are two obstacles that stand in th way of evaluating this integral. (1) The potentially non-integer power $rm/2$. (2) The integration contour is not closed in the complex plane. The first obstacle is easily remedied with an Euler substitution, $x=2t/(1-t^2)$, which renders the integrand rational. The second is overcome by representing the integral as encircling the branch cut of the integrand multiplied by a $\log$, whose jump across the branch cut is a constant.

More precisely, these manipulation give \begin{align} I(r,s,m) &= \int^{\infty}_{0}x^{r +s- 1}(1 + x)^{-s}(1 + x^2)^{-\frac{rm}{2}} dx \\ &= \int_0^1 2\frac{(1+t^2)}{(1-t^2)^2} \left(\frac{2t}{1-t^2}\right)^{r+s-1} \left(\frac{1+2t-t^2}{1-t^2}\right)^{-s} \left(\frac{1+t^2}{1-t^2}\right)^{-rm} dt \\ &= \int_0^1 2^{r+s} \frac{t^{r+s-1} (1-t^2)^{(m-1)r-1}}{(1+2t-t^2)^s (1+t^2)^{rm-1}} dt \\ &= \frac{1}{2\pi i} \oint 2^{r+s} \frac{z^{r+s-1} (1-z^2)^{(m-1)r-1}}{(1+2z-z^2)^s (1+z^2)^{rm-1}} \log[(z-1)/z] dz , \end{align} where the principal branch of $\log[(z-1)/z]$ is used and the integration contour encircles the branch cut $[0,1]$ clockwise. Deforming the contour, we can equally say that $I(r,s,m)$ is equal to the same integral, but with the contour encircling counter-clockwise each of the poles, located at $z=\pm i, 1\pm\sqrt{2}$: \begin{equation} I(r,s,m) = \sum_{w=\pm i,1\pm\sqrt{2}} \operatorname{Res}_{z=w} 2^{r+s} \frac{z^{r+s-1} (1-z^2)^{(m-1)r-1}}{(1+2z-z^2)^s (1+z^2)^{rm-1}} \log[(z-1)/z] . \end{equation} It is also conceivable that the above expression has a nice generating function, related to the one computed by Robert Israel. But I'm leaving that computation to another time.

Update: Here's part of the calculation of the generating function. It is convenient to reparametrize the indices as $k=mr-1$, so that $(m-1)r-1 = k + r$. Define $J(r,s,k)$ such that $I(r,s,m) = J(r,s,mr-1)$. The generating function for $J$ can now be written as the contour integral \begin{align} \hat{J}(u,v,w) &= \sum_{q,s,k=0}^\infty u^s v^q w^k J(q,s,k) \\ &= \sum_{q,s,k=0}^\infty \frac{1}{2\pi i}\oint \left[ u \frac{2 z}{1 + 2 z - z^2} \right]^s \left[ v \frac{2 z}{1 - z^2} \right]^{q} \left[ w \frac{1 - z^2}{1 + z^2} \right]^k \frac{\log[(z-1)/z]}{z} dz \\ &= \frac{1}{2\pi i} \oint \frac{(z^2 - 2 z - 1) (z^4 - 1)}{([z - (1 - u)]^2 - [ 1 + (1 - u)^2]) ([z + v]^2 - [1 + v^2]) (z^2 - [-\frac{1 - w}{1 + w}])} \frac{\log[(z-1)/z]}{z(1+w)} dz , \end{align} where the contour is similar to the one before, clockwise enclosing the $[0,1]$ branch cut. However, to ensure the ability to exchange integration and summation, the contour has to be first deformed into one that is confined to an annular domain that excludes neighborhood of the cut $[0,1]$ and of the poles at $z=\pm i,1\pm\sqrt{2}$. Each sum then converges absolutely and uniformly on this annular domain for sufficiently small $|u|,|v|,|w|<1$. After the summation, the contour can now be deformed to counter-clockwise enclose the poles of the integrand at $z=\pm\sqrt{-\frac{1-w}{1+w}},1-u\pm\sqrt{1+(1-u)^2},-\sqrt{1+v^2}-v$, each of which is simple. Note that the other poles are not included because they are on the other side of the contour that was held fixed during the summation. The challenge now is to express the sum of the residues over these poles in a nice way.

Update: Summing over the poles mentioned above, involves pairs of poles that are roots of quadratic polynomials with quadratic coefficients. Summation over such pairs can be expressed in a way that minimizes the use of quadratic surds. Suppose that each such root can be written as $z=x\pm y$, where $y=\sqrt{y^2}$ and $y^2$ is rational. Then (with subscripts denoting odd or even parts in $y$) \begin{align} \sum_{\alpha=x\pm y} \operatorname{Res}_{z=\alpha} \frac{f(z)}{([z-x]^2-y^2)} \log[(z-1)/z] &= \sum_\pm \frac{f(x\pm y)}{\pm 2 y} \log\frac{x\pm y-1}{x\pm y} \\ &= \left( \frac{f(x+y)}{y}\log\frac{x+y-1}{x+y} \right)_{even} \\ &= \frac{f(x+y)_{odd}}{y}\frac{1}{2}\log\left|\frac{(x-1)^2-y^2}{x^2-y^2}\right| \\ &\quad{} + f(x+y)_{even} \frac{1}{2y}\log\left|\frac{(x-1+y)(x-y)}{(x-1-y)(x+y)}\right| \end{align} From the above expression, it is obvious, that the surd $y$ will only appear in the rational form $y^2$ in all cases except the last logarithm. The same trick cannot be used for the unpaired pole $z=-\sqrt{1+v^2}-v$, so the surd $y=\sqrt{1+v^2}$ will appear explicitly in the corresponding residue. (An earlier try of mine resulted in a generating function consisting of a sum of terms, each of which had some negative powers of $u,v,w$ that would ultimately cancel. I thought it would be more convenient to find an expression where each term had a normal Taylor, rather than Laurent, expansion.)

Putting all these tricks together and using computer algebra, an explicit formula for $\hat{J}(u,v,w)$ is given by multiplying the first two entries in each row and summing over the rows in the table below: \begin{equation} \begin{array}{cccc} & \log & x & y\\ \hline \frac{( u - 1 )^2 u w}{2 ( v - u + 1 ) ( - u^2 w^2 + 2 u w^2 - w^2 + u^2 - 2 u + 2 )} & \frac{\log \left( 2 ( 1 - u ) \right)}{2} & 1 - u & \sqrt{1 + ( 1 - u)^2} \\ \frac{( 1 - u ) u ( u^2 - 2 u + 2 )}{2 ( v - u + 1 ) ( - u^2 w^2 + 2 u w^2 - w^2 + u^2 - 2 u + 2 )} & \frac{\log \left( \frac{( y - 1 + u ) ( y - u ) }{( y + 1 - u ) ( y + u)} \right) }{2 y} & 1 - u & \sqrt{1 + ( 1 - u)^2} \\ \frac{w ( u v w^2 + u w^2 - w^2 - u v - u + 2 )}{2 ( w - 1 ) ( w + 1 ) ( u^2 w^2 - 2 u w^2 + w^2 - u^2 + 2 u - 2 ) ( v^2 w^2 - v^2 - 1 )} & \frac{\log \left( \frac{2}{1 - w} \right)}{2} & 0 & \sqrt{- \frac{1 - w}{1 + w}} \\ \frac{w ( u v w^2 - v w^2 - u v + 2 v + u)}{2 ( w + 1 ) ( u^2 w^2 - 2 u w^2 + w^2 - u^2 + 2 u - 2 ) ( v^2 w^2 - v^2 - 1 )} & \frac{\log \left( \frac{1 - y}{1 + y} \right)}{2 y} & 0 & \sqrt{- \frac{1 - w}{1 + w}} \\ \frac{v ( v + 1 ) ( 2 y - v^2 - 2 )}{( v - u + 1 ) ( 2 v w y - v^2 y - 2 y - v^3 w - 2 v w + 2 v^2 + 2 )} & \log\left( \frac{2 v}{y - 1 + v} \right) & - v & \sqrt{1 + v^2} \end{array} \end{equation}

This was a fun exercise in computing contour integrals and generating functions. Don't know if it's of any use to the OP though, since this formula does not translate into a direct formula for the coefficients of the generating function.

Answer 4

Let $I(a,b,m) = \displaystyle \int_0^{\infty} \dfrac{x^{a+b-1} dx}{(1+x)^{b}(1+x^2)^{am/2}}$.

Setting $x = \tan(\theta)$. We then get that

$$I(a,b,m) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \sec^2(\theta) d \theta}{(1+\tan(\theta))^b \left(1+\tan^2(\theta) \right)^{am/2}}$$

$$I(a,b,m) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \sec^2(\theta) d \theta}{(1+\tan(\theta))^b \sec^{am}(\theta)}$$

$$I(a,b,m) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \cos^{am-2}(\theta) d \theta}{(1+\tan(\theta))^b}$$

Let us use the following short hand notation. $c = \cos(\theta)$ and $s = \sin(\theta)$. We then get that $$I(a,b,m) = \int_0^{\pi/2} \dfrac{s^{a+b-1} c^{am-a-1} d \theta}{(s+c)^b}$$ Hence, we are interested in evaluating integral of the form $$J(p,q,r) = \int_0^{\pi/2} \dfrac{s^p c^q d \theta}{(s+c)^r}$$ First some observations: $$J(p,q,r) = J(q,p,r) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\star)$$ $$J(p,q,0) = \dfrac{\beta((p+1)/2,(q+1)/2)}2$$ $$J(p,q,-1) = \int_0^{\pi/2} s^p c^q (s+c) d \theta = J(p+1,q,0) + J(p,q+1,0)$$ Further $$J(p,q,r+2) = \int_0^{\pi/2} \dfrac{s^p c^q d \theta}{(s+c)^r (1+2sc)}$$ $$J(p,q,r+2) = \sum_{k=0}^{\infty} \int_0^{\pi/2} \dfrac{s^p c^q (-2sc)^k d \theta}{(s+c)^r}$$ $$J(p,q,r+2) = \sum_{k=0}^{\infty}(-2)^k J(p+k,q+k,r)$$

Hence, you can use this to compute $J(p,q,r)$ for all $r$ since these can be obtained as we know the values for $J(p,q,0)$ and $J(p,q,-1)$. (For even $r$, $J(p,q,r)$ will eventually depend on $J(p,q,0)$ and for odd $r$, $J(p,q,r)$ will eventually depend on $J(p,q,-1)$)

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Few other relations, which might be of use. We have $$\lim_{p \to \infty} J(p,q,r) = \lim_{q \to \infty} J(p,q,r) = \lim_{r \to \infty} J(p,q,r) = 0$$ Note that $$J(p+2,q,r) = \int_0^{\pi/2} \dfrac{s^p c^q (1-c^2) d \theta}{(s+c)^r}$$ Hence, we get that $$J(p+2,q,r) + J(p,q+2,r) = J(p,q,r)$$ Setting $p=q$, and making use of $(\star)$, we get that $$J(p+2,p,r) = \dfrac{J(p,p,r)}2$$ Note that $$J(0,0,r) = \int_0^{\pi/2} \dfrac{d \theta}{(s+c)^r} = \int_0^{\pi/2} \dfrac1{2^{r/2}} \dfrac{d \theta}{\sin^r(\theta + \pi/4)}$$

Answer 5

Not quite an answer, but: If you call the integrand $f(x, r, s, m),$ and evaluate for explicit values of $r, s, m$ you get a clue. For example, if you define $g(x, k) = f(x,1, 2, k),$ then $F(k)=\int_0^\infty g(x, k) dx,$ for odd $k$ seems to equal $F(k)=\frac{2^{(k-1)/2} G_{3,3}^{3,3}\left(1\left| \begin{array}{c} -\frac{1}{2},-\frac{1}{2},0 \\ 0,\frac{1}{2},(k-3)/2 \\ \end{array} \right.\right)}{(k-2)!! \pi ^{3/2}}$

Where the $G$ is the Meijer G function, and $n!!$ is the product of all odd numbers not exceeding $n.$

For even $k$ the answer is always a linear function of $\pi$ with rational coefficients, e.g. $F(4) = \frac12 - \frac{\pi}8$

$F(6)= \frac14 - \frac{\pi}{16}$

$F(8) = \frac1{12} - \frac{\pi}{64}$

$F(10)= \frac{\pi}{128}$

$F(12) =-\frac{1}{30}+ \frac{17\pi}{1024}$

And so on. I am not sure I am quite catching the pattern of the coefficients (the powers of two in the denominator are not quite regular, even).

I did not try the other $r, s$ but this already seems pretty interesting.