# Closed-form expression for this definite integral

Does this definite integral have a closed-form expression?

\begin{align*} I &= \int_0^\infty \sqrt{ \frac{1}{2} \frac{1}{x} \left( \frac{1}{(1+x)^2} + \frac{z}{(1+xz)^2} \right) } \, dx \\ &= \frac{1}{\sqrt{2}} \int_0^\infty \frac{1}{x} \sqrt{ \frac{x}{1+x} \left( 1-\frac{x}{1+x} \right) + \frac{xz}{1+xz} \left( 1-\frac{xz}{1+xz} \right) } \, dx \end{align*}

where $x>=0$, and $z>=0$ is constant.

I have tried the following substitution:

$$u = \frac{xz}{1+xz}, \quad x = \frac{1}{z} \frac{u}{(1-u)}, \quad dx = \frac{1}{z} \frac{1}{(1-u)^2} du$$

$$I = \frac{1}{\sqrt{2}} \int_0^1 \frac{1}{\sqrt{u(1-u)}} \sqrt{ 1 + \frac{z}{((1-u)z+u)^2} } \, du$$

...but I can't seem to make any more progress. Using $u=\frac{x}{1+x}$ yields a similar expression which is also difficult to simplify.

Can you find a way to simplify this further?

Update (Nov 4, 2013): The integral can be rearranged to the following form, which might make it easier to match something from a table of known integrals:

$$I = \frac{1}{\sqrt{2}} \sqrt{\frac{1{+}z}{z}} \int_0^1 u^{-\frac{1}{2}} (1{-}u)^{-\frac{1}{2}} \left( 1+\frac{1{-}z}{z} u \right)^{-1} \left( 1 + 2\frac{1{-}z}{1{+}z}u + \frac{(1{-}z)^2}{z(1{+}z)}u^2 \right)^{\frac{1}{2}} \, du$$

• What is $z$? A constant? – Clayton Aug 30 '13 at 15:17
• If it were minus instead of plus in the square root then the integral would be very doable using residues. – Ron Gordon Aug 30 '13 at 15:19
• @Clayton: Yes, $z$ is a constant. I updated the question to make this explicit. – Tyler Streeter Aug 30 '13 at 15:23

$${\rm I}\left(z\right) \equiv \int_{0}^{\infty}\sqrt{\vphantom{\LARGE A^{A}}\frac{1}{2x} \left\lbrack% \frac{1}{\left(1 + x\right)^{2}} + \frac{z}{\left(1 + xz\right)^2} \right\rbrack\,}\ {\rm d}x\,, \qquad \begin{array}{|rclcl} \,\,{\rm I}\left(0\right) & = & {\rm I}\left(\infty\right) & = & {\sqrt{2\,} \over 2}\,\pi \\[1mm] \,\,{\rm I}\left(1\right) & = & \pi&& \end{array}$$

\begin{align} {\rm I}\left(z\right) &= {\sqrt{2} \over 2}\int_{0}^{\infty}\!\! {\sqrt{z\left(z + 1\right)x^{2} + 4zx + 1 + z\,} \over \left(x + 1\right)\left(1 + xz\right)}\, {{\rm d}x \over \sqrt{x\,}} \\[3mm]&= {\sqrt{2} \over 2}\left(z + 1 \over z\right)^{1/2}\int_{0}^{\infty} {\sqrt{x^{2} + 4\left(z + 1\right)^{-1}\,x + z^{-1}\,} \over \left(x + 1\right)\left(x + z^{-1}\right)} {{\rm d}x \over \sqrt{x\,}}& \end{align}

Equation $x^{2} + 4\left(z + 1\right)^{-1}\,x + z^{-1} = 0$ has the complex roots:

$$x_{\pm} = -\,{2 \over z + 1} \pm {\rm i}\,{\left\vert z -1\right\vert \over \left(z + 1\right)\,\sqrt{z\,}}\,, \qquad\mbox{Notice that}\quad x_{+}x_{-} = \left\vert x_{\pm}\right\vert^{2} = z^{-1}$$

Then, ${\rm I}\left(z\right) = \left(\sqrt{2}/2\right)\sqrt{a^{2} + 1\,}\ {\cal I}\left(a\right)$ where $a = z^{-1/2}$.

\begin{eqnarray*} {\cal I}\left(a\right) & \equiv & \int_{0}^{\infty} {\sqrt{x^{2} + 4a^{2}\left(a^{2} + 1\right)^{-1}\,x + a^{2}\,} \over \left(x + 1\right)\left(x + a^{2}\right)} {{\rm d}x \over \sqrt{x\,}} \\ x_{\pm} & = & {-2 \pm {\rm i}\,\left\vert\,a^{2} - 1\right\vert \over a^{2} + 1}\,a^{2} \end{eqnarray*}

$${\rm I}\left(z\right) = {\sqrt{2\,} \over 2}\,\left(z + 1 \over z\right)^{1/2}{\cal I}\left(1 \over \sqrt{z\,}\right)\,, \qquad\qquad {\cal I}\left(a\right) = {\sqrt{2\,} \over \sqrt{a^{2} + 1\,}}\,{\rm I}\left(1 \over a^{2}\right)$$

$${\cal I}\left(a\right) = \int_{-\infty}^{\infty} {\sqrt{x^{4} + 4a^{2}\left(a^{2} + 1\right)^{-1}\,x^{2} + a^{2}\,} \over \left(x^{2} + 1\right)\left(x^{2} + a^{2}\right)} \,{\rm d}x$$

Integration over the complex plane is possible. You have to take into account that $$x^{4} + 4a^{2}\left(a^{2} + 1\right)^{-1}\,x^{2} + a^{2} = \left(x - x_{\atop -}^{1/2}\right)\left(x + x_{\atop -}^{1/2}\right) \left(x - x_{\atop +}^{1/2}\right)\left(x + x_{\atop +}^{1/2}\right)$$ which introduces branch-cuts in the complex plane.

• Thanks for the answer. I'm not too familiar with integrals in the complex plane, so I will read more about that. Also, note that I updated the question to make it explicit that $x>=0$, in case that matters. – Tyler Streeter Sep 6 '13 at 12:25
• @TylerStreeter You're welcome. – Felix Marin Sep 6 '13 at 23:41