1
$\begingroup$

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?

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

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 $$ u = \tanh \frac{1}{2} x = \frac{\sinh x}{\cosh x + 1} = \frac{\cosh x - 1}{\sinh x} $$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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 $\cosh \alpha = \sqrt{5}$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ very nice. .... $\endgroup$ – Will Jagy Jan 23 '18 at 23:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.