I need to solve the following integral: $$ I = \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x}. $$ Wolfram Alpha gives the answer as $ \dfrac{\pi}{2 \sqrt{2}} $.

I think it’s achievable by complex analysis, but I really have no idea how. Also, is there a special name for this integral, i.e., does it have some known physical significance?

  • 1
    $\begingroup$ The integrand has a closed-form antiderivative. $\endgroup$ Nov 13, 2015 at 2:55
  • $\begingroup$ You don't need to use complex analysis. $\endgroup$
    – Ben Longo
    Nov 13, 2015 at 3:03
  • $\begingroup$ if you want to use contour integration, take a keyhole around with a slit at the line $(1,\infty)$ $\endgroup$
    – tired
    Nov 13, 2015 at 10:30

6 Answers 6


Let $u=\sqrt{x-1}$, $du=\frac{1}{2\sqrt{x-1}}$, then


We can apply partial fractions here.



The first integrand is almost $\tan^{-1}$, so we can factor the 2 and apply the substitution $s=\frac{u}{\sqrt{2}}$.

$$I=\sqrt 2\int_0^\infty\frac{1}{s^2+1}ds-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$

$$=\frac{\pi}{\sqrt 2}-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$

Now to tackle the second integral we can use a trig sub.

Let $u=\sqrt{2}\tan (t)$, $du=\sqrt 2 \sec^2(t)dt$.

$$I=\frac{\pi}{\sqrt 2}-4\sqrt 2 \int_0^{\pi/2}\frac{\sec ^2 t}{4\sec^4t}dt$$

$$=\frac{\pi}{\sqrt 2}-\sqrt 2\int _0^{\pi/2}\cos^2(t) dt$$

$$=\frac{\pi}{\sqrt 2}-\sqrt 2 \int_0^{\pi/2}\left(\frac{1}{2}\cos(2t)+\frac{1}{2} \right)dt$$

If we substitute $v=2t$ and split the integral up we get

$$I=\frac{\pi}{\sqrt 2}-\frac{1}{2\sqrt 2} \int_0^\pi \cos(v)dv-\frac{1}{\sqrt 2}\int_0^{\pi/2}1dt$$

The first integral clearly goes to 0 and the second integral becomes $\frac{\pi}{2\sqrt 2}$.


$$I=\frac{\pi}{2\sqrt 2}$$

  • 1
    $\begingroup$ There's a much easier way to find the $u$ integral. Using IBP gives $$\int u\cdot\frac{2u}{(u^2+2)^2}du = -u\cdot\frac{1}{u^2+2} + \int \frac{1}{u^2+2}\,du = -\frac{u}{u^2+2} + \arctan u$$ $\endgroup$
    – Dylan
    Nov 18, 2015 at 5:51

Using a keyhole countour with the origin of the keyhole at $z=1$ and the small circle enclosing the value $z=1$ and using $$f(z) = \frac{\exp(1/2\log(z-1))}{(1+z)^2}$$ with the branch cut of the logarithm on the positive real axis and the argument from $0$ to $2\pi$ we get for the integral $$I=\int_1^\infty \frac{\sqrt{x-1}}{(1+x)^2} dx$$ that $$I(1-\exp(\pi i)) = 2\pi i\mathrm{Res}_{z=-1} f(z)$$ or $$I = \pi i\mathrm{Res}_{z=-1} f(z).$$

Now the logarithmic term is certainly analytic in a neighborhood of $z=-1$ and we have $$\mathrm{Res}_{z=-1} f(z) = \left.\left(\exp(1/2\log(z-1))\right)'\right|_{z=-1} = \left. \left(\exp(1/2\log(z-1))\right) \frac{1}{2}\frac{1}{z-1}\right|_{z=-1} \\= \exp(1/2(\log 2 + \pi i)) \times -\frac{1}{2}\frac{1}{2} = -\frac{1}{4} \sqrt{2} i.$$

This yields $$I = -\frac{1}{4} \sqrt{2} i \times\pi i=\frac{\sqrt{2}\pi}{4}.$$

Remark. The estimates for the circular components are done using ML same as at this MSE link. We get for the large circle parameterized by $z=R e^{it}$ $$\lim_{R\rightarrow \infty} 2\pi R \frac{\sqrt{R+1}}{(R-1)^2} = 0.$$

The small circle is a parameterized with $z=1+\epsilon e^{it}$ and we get $$\lim_{\epsilon\rightarrow 0} 2\pi\epsilon \frac{\sqrt{\epsilon}}{4} = 0.$$


$$I=\int_{1}^{\infty}\frac{\sqrt{x-1}}{(x+1)^2}dx$$ Use substitution $\frac{1}{x+1}=t$ which implies $\frac{dx}{(x+1)^2}=-dt$ So


Again use substitution $\sqrt{t}=y$ which implies $\frac{dt}{\sqrt{t}}=2dy$


Use standard integral $$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}(\frac{x}{a})$$

we get $$I=\frac{\pi}{2\sqrt{2}}$$


Hint Substitution $x=t^2+1$...


This is the general antiderivative. Just take the limits and you're good. $$\int{\frac{\sqrt{x-1}}{(x+1)^2}dx}$$ Substitute $u = \sqrt{x-1}$ $$= \int{\frac{u}{(u^2+2)^2}du}$$ $$= 2\int\left(\frac{u}{u^2+2}-\frac{2}{(u^2+2)^2}\right)du$$ $$= \int\frac{u}{\frac{u^2}{2}+1}du-4\int\frac{2}{(u^2+2)^2}du$$ Substitute $s = \frac{u}{\sqrt{2}}$ and $ds = \frac{1}{\sqrt{2}}$ $$= \sqrt{2}\int\frac{1}{s^2+1}ds-4\int\frac{2}{(u^2+2)^2}du$$ $$= \sqrt{2}\arctan(s)-4\int\frac{2}{(u^2+2)^2}du$$ Substitute $u = \sqrt{2}\tan (p)\quad$ and $\quad du = \sqrt{2}\sec^2(p) dp\quad$ and $\quad(u^2 + 2)^2 = (2\tan^2(p)+2)^2 = 4\sec^4(p)\quad$ and $\quad p=\arctan\frac{u}{\sqrt{2}}$ $$= \sqrt{2}\arctan(s)-\sqrt{2}\int\cos^2(p)du$$ $$= \sqrt{2}\arctan(s)-\sqrt{2}\int\left(\frac{1}{2} \cos(2p) + \frac{1}{2}\right)du$$ $$= \sqrt{2}\arctan(s)-\frac{1}{\sqrt{2}}\int\cos(2p) - \frac{1}{\sqrt{2}}\int du$$ $$= \sqrt{2}\arctan(s)-\frac{p}{\sqrt{2}} - \frac{\sin(p)\cos(p)}{\sqrt{2}}$$ $$= \sqrt{2}\arctan(s)-\frac{\arctan\frac{u}{\sqrt{2}}}{\sqrt{2}} - \frac{\sin(\arctan\frac{u}{\sqrt{2}})\cos(\arctan\frac{u}{\sqrt{2}})}{\sqrt{2}}$$ Note that $\cos(\arctan(z)) = \frac{1}{\sqrt{z^2 + 1}}$ and $\sin(\arctan(z)) = \frac{z}{\sqrt{z^2+1}}$ $$= \frac{2\sqrt{2}(u^2+2)\arctan(s)+\sqrt{2}(u^2+2)\arctan\frac{u}{\sqrt{2}}+2u}{2(u^2+2)}$$ $$=\frac{\sqrt{2}(u^2+2)\arctan \frac{u}{\sqrt{2}}-2u}{2(u^2+2)}$$ $$=\frac{\sqrt{2}(x+1)\arctan \frac{\sqrt{x-1}}{\sqrt{2}}-2\sqrt{x-1}}{2(x+1)}$$ $$=\frac{\arctan \frac{\sqrt{x-1}}{\sqrt{2}}}{\sqrt{2}}-\frac{\sqrt{x-1}}{x+1}$$


I think I have a different answer.

Let $\frac{x+1}{2} = u$, so that $dx = 2 du$ and $x-1 = 2(u-1)$. Then

$$\begin{align} \int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx &= \int_1^\infty \frac{\sqrt{2} (u-1)^{\frac{1}{2}}}{4u^2} 2 du \\ &= \frac{1}{\sqrt{2}} \int_1^\infty u^{-2} (u-1)^\frac{1}{2} du \end{align}$$ Now let $u = \frac{1}{t}$, so that $du = \frac{-1}{t^2} dt$. Continuing. $$\begin{align} \phantom{\int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx} &= \frac{1}{\sqrt{2}} \int_0^1 t^2 (1-t)^\frac{1}{2} \frac{1}{\sqrt{t}} \frac{1}{t^2} dt \\ &= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{-1}{2}} (1-t)^\frac{1}{2} dt \\ &= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{1}{2}-1} (1-t)^{\frac{3}{2} - 1} dt \\ &= \frac{1}{\sqrt{2}} B ( \frac{1}{2} . \frac{3}{2} ) \\ &= \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{3}{2})} {\sqrt{2}\Gamma(2)} \\ &= \frac{\pi}{2\sqrt{2}}. \end{align}$$

Original image of work by hand.

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    $\begingroup$ Honestly, with 1500 points I expect you know the customs here, and I see many answers where you used LaTeX. You could make an effort and write your answer correctly. $\endgroup$ Nov 14, 2015 at 0:48

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