# A faster way to evaluate $\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt$?

I need to evaluate the integral

$$\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt\tag1$$

After some workarounds I found the change of variable $t=2\sqrt{x^2-1}$, then

$$\int_1^\infty\frac{2\sqrt{1+(t/2)^2}}{t^3}\,\mathrm dt=\frac12\int_{\sqrt5/2}^\infty\frac{x^2}{(x^2-1)^2}\,\mathrm dx\\=\frac12\left[\frac{x}{2(1-x^2)}\bigg|_{\sqrt5/2}^\infty+\frac12\int_{\sqrt5/2}^\infty\frac{\mathrm dx}{x^2-1}\right]\\=\frac{\sqrt5}2+\frac18\int_{\sqrt5/2}^\infty\left(\frac1{x-1}-\frac1{x+1}\right)\,\mathrm dx\\=\frac{\sqrt5}2+\frac18\ln\left(\frac{\sqrt 5+2}{\sqrt 5-2}\right)\\=\frac{\sqrt5}2+\frac14\ln(\sqrt5+2)$$

But my intuition says that it must exists a more straightforward way to evaluate this integral. In fact, using Wolfram Mathematica, I get the equivalent[*] result

$$\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt=\frac14(2\sqrt5+\operatorname{arsinh}(2))$$

[*] The equivalence can be seen from

$$\operatorname{arsinh}(x)=\ln(x+\sqrt{1+x^2})$$

My question: someone knows a faster way to evaluate manually this integral? Maybe a better change of variable?

• begin with $t=2 \sinh w$ – Will Jagy Jan 23 '18 at 21:48
• @WillJagy I did it, but this doesnt seem a big improvement. You find from it something like $\int \frac1{\sinh x\tanh^2 x}=\int(\sinh^{-3}x+\sinh^{-1}x)$ – Masacroso Jan 23 '18 at 22:23
• en.wikipedia.org/wiki/… – Will Jagy Jan 23 '18 at 22:44

## 2 Answers

I am very fond of hyperbolic functions for integrals. If we begin with your $t = 2 \sinh x,$ we expect to get to something consistent with the wikipedia way of writing the Weierstrass substitution for hyperbolic functions, give me a few more minutes.

$$\int \frac{\cosh^2 x}{ 2 \sinh^3 x} dx.$$ Then let us use a letter different from your $t,$ $$\sinh x = \frac{2u}{1 - u^2}, \; \; \; \frac{1}{\sinh x} = \frac{1 - u^2}{2u}$$ $$\cosh x = \frac{1 + u^2}{1 - u^2},$$ $$d x = \frac{2du}{1 - u^2} \; .$$ $$\int \frac{(1 + u^2)^2 (1 - u^2)^3 2 du}{2 (1 - u^2)^2 (2u)^3 (1-u^2)}$$ $$\int \frac{(1 + u^2)^2 du}{ (2u)^3}$$ $$\int \frac{1 + 2u^2 + u^4 }{ 8u^3} \; du$$ $$\int \frac{1}{8u^3} + \frac{1}{4u} + \frac{u}{8} \; \; du$$

They give more detail here, and we do need an expression for our $u$ That comes out $$u = \tanh \frac{1}{2} x = \frac{\sinh x}{\cosh x + 1} = \frac{\cosh x - 1}{\sinh x}$$

Enforcing a substitution of $x \mapsto 1/x$ we have $$I = \int_0^1 \sqrt{4 x^2 + 1} \, dx.$$ This integral can be readily found using a hyperbolic substitution of $x \mapsto \frac{1}{2} \sinh x$. Doing so yields \begin{align*} I &= \frac{1}{2} \int_0^{\sinh^{-1} (2)} \cosh^2 x \, dx\\ &= \frac{1}{4} \int_0^{\sinh^{-1} (2)} [\cosh (2x) + 1] \, du \tag1\\ &= \frac{1}{4} \left [\frac{1}{2} \sinh (2x) + x \right ]_0^{\sinh^{-1} (2)}\\ &= \frac{1}{8} \sinh [2\sinh^{-1} (2)] + \frac{1}{4} \sinh^{-1} (2)\\ &= \frac{\sqrt{5}}{2} + \frac{1}{4} \sinh^{-1} (2) \tag2 \end{align*}

Explanation

(1) Using $\cosh^2 x = \frac{1}{2} (\cosh (2x) + 1)$

(2) Using $\sinh (2 \alpha) = 2 \sinh \alpha \cosh \alpha$ where $\alpha = \sinh^{-1} (2)$ such that $\sinh \alpha = 2$ and $\cosh \alpha = \sqrt{5}$

• very nice. .... – Will Jagy Jan 23 '18 at 23:19