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Hi guys I have the following definite integral to solve:

$$\int_{0}^{\infty}e^{-u}\frac{1}{\left(\sqrt{1+(h+u)^{2}}\right)^{5}}du$$

is it possible to obtain an analytic expression? And if not why?

When I try the integration in mathematica it gives me back the integrand...

thank you in advance

Edit: $h$ is a real number so that $ 1 + (h+u)^{2}$ is always positive

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  • $\begingroup$ I believe that if you view the integral as a function $I(h)$ of $h$, you can evaluate $I'(h)$ by differentiating inside the integral and then using integration by parts. $\endgroup$ – Aaron Aug 25 '14 at 16:53
  • $\begingroup$ Could you post the first steps in an answer so ti make clear What you mean..? $\endgroup$ – SSC Napoli Aug 25 '14 at 16:57
  • $\begingroup$ I wrote up a 'solution' based on my comment. It doesn't quite get what we want, but it pairs nicely with the answer by @Lucian. $\endgroup$ – Aaron Aug 25 '14 at 17:49
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$\int_0^\infty e^{-u}\dfrac{1}{\left(\sqrt{1+(h+u)^2}\right)^5}du$

$=\int_0^\infty\dfrac{e^{-u}}{(1+(u+h)^2)^\frac{5}{2}}du$

$=\int_h^\infty\dfrac{e^{-(u-h)}}{(1+u^2)^\frac{5}{2}}d(u-h)$

$=e^h\int_h^\infty\dfrac{e^{-u}}{(1+u^2)^\frac{5}{2}}du$

Consider $\int\dfrac{1}{(1+u^2)^\frac{5}{2}}du$ :

Let $u=\tan\theta$ ,

Then $du=\sec^2\theta~d\theta$

$\therefore\int\dfrac{1}{(1+u^2)^\frac{5}{2}}du$

$=\int\dfrac{\sec^2\theta}{(1+\tan^2\theta)^\frac{5}{2}}d\theta$

$=\int\dfrac{\sec^2\theta}{\sec^5\theta}d\theta$

$=\int\cos^3\theta~d\theta$

$=\int\cos^2\theta~d(\sin\theta)$

$=\int(1-\sin^2\theta)~d(\sin\theta)$

$=\sin\theta-\dfrac{\sin^3\theta}{3}+C$

$=\dfrac{u}{\sqrt{1+u^2}}-\dfrac{u^3}{3(1+u^2)^\frac{3}{2}}+C$

$=\dfrac{3u(1+u^2)-u^3}{3(1+u^2)^\frac{3}{2}}+C$

$=\dfrac{3u+2u^3}{3(1+u^2)^\frac{3}{2}}+C$

$=\dfrac{2u(1+u^2)+u}{3(1+u^2)^\frac{3}{2}}+C$

$=\dfrac{2u}{3\sqrt{1+u^2}}+\dfrac{u}{3(1+u^2)^\frac{3}{2}}+C$

$\therefore e^h\int_h^\infty\dfrac{e^{-u}}{(1+u^2)^\frac{5}{2}}du$

$=e^h\int_h^\infty e^{-u}~d\left(\dfrac{2u}{3\sqrt{1+u^2}}+\dfrac{u}{3(1+u^2)^\frac{3}{2}}\right)$

$=e^h\left[e^{-u}\left(\dfrac{2u}{3\sqrt{1+u^2}}+\dfrac{u}{3(1+u^2)^\frac{3}{2}}\right)\right]_h^\infty-e^h\int_h^\infty\left(\dfrac{2u}{3\sqrt{1+u^2}}+\dfrac{u}{3(1+u^2)^\frac{3}{2}}\right)d(e^{-u})$

$=-\dfrac{2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+e^h\int_h^\infty e^{-u}\left(\dfrac{2u}{3\sqrt{1+u^2}}+\dfrac{u}{3(1+u^2)^\frac{3}{2}}\right)du$

$=-\dfrac{2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+e^h\int_h^\infty\dfrac{2ue^{-u}}{3\sqrt{1+u^2}}du+e^h\int_h^\infty\dfrac{ue^{-u}}{3(1+u^2)^\frac{3}{2}}du$

$=-\dfrac{2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+e^h\int_h^\infty\dfrac{2ue^{-u}}{3\sqrt{1+u^2}}du+e^h\int_h^\infty\dfrac{e^{-u}}{6(1+u^2)^\frac{3}{2}}d(1+u^2)$

$=-\dfrac{2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+e^h\int_h^\infty\dfrac{2ue^{-u}}{3\sqrt{1+u^2}}du-e^h\int_h^\infty\dfrac{e^{-u}}{3}d\left(\dfrac{1}{\sqrt{1+u^2}}\right)$

$=-\dfrac{2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+e^h\int_h^\infty\dfrac{2ue^{-u}}{3\sqrt{1+u^2}}du-e^h\left[\dfrac{e^{-u}}{3\sqrt{1+u^2}}\right]_h^\infty+e^h\int_h^\infty\dfrac{1}{3\sqrt{1+u^2}}d(e^{-u})$

$=\dfrac{1}{3\sqrt{1+h^2}}-\dfrac{2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+\dfrac{2e^h}{3}\int_h^\infty\dfrac{ue^{-u}}{\sqrt{1+u^2}}du-\dfrac{e^h}{3}\int_h^\infty\dfrac{e^{-u}}{\sqrt{1+u^2}}du$

$=\dfrac{1-2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+\dfrac{2e^h}{3}\int_{\sinh^{-1}h}^\infty\dfrac{e^{-\sinh u}\sinh u}{\sqrt{1+\sinh^2u}}d(\sinh u)-\dfrac{e^h}{3}\int_{\sinh^{-1}h}^\infty\dfrac{e^{-\sinh u}}{\sqrt{1+\sinh^2u}}d(\sinh u)$

$=\dfrac{1-2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+\dfrac{2e^h}{3}\int_{\sinh^{-1}h}^\infty e^{-\sinh u}\sinh u~du-\dfrac{e^h}{3}\int_{\sinh^{-1}h}^\infty e^{-\sinh u}~du$

$=\dfrac{1-2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+\dfrac{2e^h}{3}\int_{\sinh^{-1}h}^\infty e^{-\frac{e^u-e^{-u}}{2}}\dfrac{e^u-e^{-u}}{2}du-\dfrac{e^h}{3}\int_{\sinh^{-1}h}^\infty e^{-\frac{e^u-e^{-u}}{2}}du$

$=\dfrac{1-2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+\dfrac{e^h}{3}\int_{\sinh^{-1}h}^\infty e^{-\frac{e^u}{2}+\frac{e^{-u}}{2}}e^u~du-\dfrac{e^h}{3}\int_{\sinh^{-1}h}^\infty e^{-\frac{e^u}{2}+\frac{e^{-u}}{2}}du-\dfrac{e^h}{3}\int_{\sinh^{-1}h}^\infty e^{-\frac{e^u}{2}+\frac{e^{-u}}{2}}e^{-u}~du$

$=\dfrac{1-2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+\dfrac{e^h}{3}\int_{e^{\sinh^{-1}h}}^\infty e^{-\frac{u}{2}+\frac{1}{2u}}u~d(\ln u)-\dfrac{e^h}{3}\int_{e^{\sinh^{-1}h}}^\infty e^{-\frac{u}{2}+\frac{1}{2u}}d(\ln u)-\dfrac{e^h}{3}\int_{e^{\sinh^{-1}h}}^\infty\dfrac{e^{-\frac{u}{2}+\frac{1}{2u}}}{u}d(\ln u)$

$=\dfrac{1-2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+\dfrac{e^h}{3}\int_{e^{\sinh^{-1}h}}^\infty e^{-\frac{u}{2}+\frac{1}{2u}}du-\dfrac{e^h}{3}\int_{e^{\sinh^{-1}h}}^\infty\dfrac{e^{-\frac{u}{2}+\frac{1}{2u}}}{u}du-\dfrac{e^h}{3}\int_{e^{\sinh^{-1}h}}^\infty\dfrac{e^{-\frac{u}{2}+\frac{1}{2u}}}{u^2}du$

$=\dfrac{1-2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+\dfrac{e^h}{3}\int_1^\infty e^{-\frac{e^{\sinh^{-1}h}u}{2}+\frac{1}{2e^{\sinh^{-1}h}u}}d(e^{\sinh^{-1}h}u)-\dfrac{e^h}{3}\int_1^\infty\dfrac{e^{-\frac{e^{\sinh^{-1}h}u}{2}+\frac{1}{2e^{\sinh^{-1}h}u}}}{e^{\sinh^{-1}h}u}d(e^{\sinh^{-1}h}u)-\dfrac{e^h}{3}\int_1^\infty\dfrac{e^{-\frac{e^{\sinh^{-1}h}u}{2}+\frac{1}{2e^{\sinh^{-1}h}u}}}{(e^{\sinh^{-1}h}u)^2}d(e^{\sinh^{-1}h}u)$

$=\dfrac{1-2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+\dfrac{e^{h+\sinh^{-1}h}}{3}\int_1^\infty e^{-\frac{e^{\sinh^{-1}h}u}{2}+\frac{e^{-\sinh^{-1}h}}{2u}}du-\dfrac{e^h}{3}\int_1^\infty\dfrac{e^{-\frac{e^{\sinh^{-1}h}u}{2}+\frac{e^{-\sinh^{-1}h}}{2u}}}{u}du-\dfrac{e^{h-\sinh^{-1}h}}{3}\int_1^\infty\dfrac{e^{-\frac{e^{\sinh^{-1}h}u}{2}+\frac{e^{-\sinh^{-1}h}}{2u}}}{u^2}du$

$=\dfrac{1-2h}{3\sqrt{1+h^2}}-\dfrac{h}{3(1+h^2)^\frac{3}{2}}+\dfrac{e^{h+\sinh^{-1}h}}{3}K_{-1}\left(\dfrac{e^{\sinh^{-1}h}}{2},-\dfrac{e^{-\sinh^{-1}h}}{2}\right)-\dfrac{e^h}{3}K_0\left(\dfrac{e^{\sinh^{-1}h}}{2},-\dfrac{e^{-\sinh^{-1}h}}{2}\right)-\dfrac{e^{h-\sinh^{-1}h}}{3}K_1\left(\dfrac{e^{\sinh^{-1}h}}{2},-\dfrac{e^{-\sinh^{-1}h}}{2}\right)$

(according to http://artax.karlin.mff.cuni.cz/r-help/library/DistributionUtils/html/incompleteBesselK.html)

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  • $\begingroup$ there's a typo in the last step, the $K$ function multiplying $+\frac{e^{h+sinh^{-1}h}}{3}$ should be $K_{1}$; While the $K$ function multiplying $-\frac{e^{h-sinh^{-1}h}}{3}$ should be $K_{-1}$ $\endgroup$ – SSC Napoli Sep 5 '14 at 18:52
  • $\begingroup$ @user3810266 Note that the incomplete bessel K function defined in artax.karlin.mff.cuni.cz/r-help/library/DistributionUtils/html/… is $K_\nu(x,y)=\int_1^\infty t^{-\nu-1}e^{-xt-\frac{y}{t}}~dt=\int_1^\infty\dfrac{e^{-xt-\frac{y}{t}}}{t^{\nu+1}}dt$ , I don't think the matching has error. $\endgroup$ – Harry Peter Sep 9 '14 at 12:15
  • $\begingroup$ ah true, I was just considering $K_{-\nu}(x,y)$; anyway i was wondering if it is possible to give a general expression for definite integrals of the kind i posted in my question, in term of the incomplete Bessel function $K_{\nu}(x,y)$ $\endgroup$ – SSC Napoli Sep 11 '14 at 11:22
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Is it possible to obtain an analytic expression?

No.

And if not, why?

Because there are no “incomplete” Bessel or Struve functions. Notice that even for the simple case

$h=0$, your integral becomes $\displaystyle\int_0^\infty\frac{e^{-x}}{\Big(\sqrt{1+x^2}\Big)^{2n+1}}dx=\frac\pi2\cdot\frac{(-1)^nH_{-n}(1)-Y_n(1)}{(2n-1)!!}$, where H

and Y are the Struve H and Bessel Y functions, respectively. However, by modifying either of the two fixed integration limits, the resulting expression cannot be parsed, even in terms of these two special functions. $\big($The standard substitution used to derive this result was $x=\sinh t\big)$.

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  • $\begingroup$ So in the simple case when h=0 there is an analytical expression in terms of H and Y of my integral? Or am i missing something? $\endgroup$ – SSC Napoli Aug 15 '14 at 1:43
  • $\begingroup$ @user3810266: Yes. Just let $n=2$ in the above expression. $\big($Also, if by any chance you are not familiar with the $!!$ notation, see double factorial$\big)$. $\endgroup$ – Lucian Aug 15 '14 at 1:48
  • $\begingroup$ Thank you for the answer, but if in my original integral i use the substitution $h+u=x$ don't I end up with the same integral posted by you? $\endgroup$ – SSC Napoli Aug 15 '14 at 1:51
  • $\begingroup$ @user3810266: No. You end up with a similar integral, whose lower integration limit if h instead of $0,$ and for which no closed form exists. $\endgroup$ – Lucian Aug 15 '14 at 2:04
  • $\begingroup$ Now it'a celar...Anyway Could you please give me a reference for this derivation..? $\endgroup$ – SSC Napoli Aug 15 '14 at 2:09
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Let us write $f(x)=(1+x^2)^{-5/2}$ so that we are trying to evaluate $I(h)=\int_0^{\infty} e^{-u}f(u+h)du$. We can differentiate $I$ with respect to $h$ by differentiating under the integral sign to get

$$ I'(h)=\int_0^{\infty} e^{-u} f'(u+h)du = [e^{-u}f(u+h)]_{u=0}^{\infty}+\int_0^{\infty} e^{-u}f(u+h)du = f(h)+I(h)$$

where we have used integration by parts with the two factors $e^{-u}$ and $f(u+h)$.

Thus, we need to solve the first order constant coefficient differential equation

$$ y'-y = f(x). $$

Multiplying through by the integrating factor $e^{-x}$ and integrating both sides, we have

$$e^{-x}y=\int e^{-x}f(x). dx$$

The good news is that we have eliminated the dependence of our integral on two variables. Unfortunately, we have added the need to find an indefinite integral. Whether this is an adequate simplification, I cannot say.

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  • $\begingroup$ a comment here, is $y=y(x)$? and once I determined $y$ how do i get the final solution to the original integral? $\endgroup$ – SSC Napoli Sep 2 '14 at 17:18
  • $\begingroup$ $y$ as a function of $x$ is the same as $I$ as a function of $h$ except that, in solving for $y$, our antiderivative is only determined up to an additive constant. To determine the proper constant, we can use either the value of $I(0)$ or the asymptotic properties of $I(\infty)$. $\endgroup$ – Aaron Sep 2 '14 at 19:00
  • $\begingroup$ unfortunately there doesn't seem to be an analytical expression for the indefinite integral...at least according to mathematica $\endgroup$ – SSC Napoli Sep 3 '14 at 15:10
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Using the substitution $t=u+h$, you get $$I=\int_0^{\infty}e^{-u}\frac{1}{\sqrt{1+(u+h)^2}^5}\,dx=e^{h}\int_h^1e^{-t}\frac{1}{\sqrt{t^2+1}^5}\,dt$$

Using the substitution $t=\tan(\theta)$, you get $dt=\sec^2(\theta)d\theta$, $t=h\rightarrow\theta=\arctan(h)=\theta_0$, $t=\infty\rightarrow\theta=\frac{\pi}{2}$:

$$\begin{array}{rcl} I=e^{h}\int_h^1e^{-t}\frac{1}{\sqrt{t^2+1}^5}\,dt&=&e^h\int_0^{\frac{\pi}{2}}e^{\tan{\theta}}\frac{\sec^2(\theta)}{\sqrt{\tan^2(\theta)+1}^5}\,d\theta\\ &=&e^h\int_{\theta_0}^{\frac{\pi}{2}}e^{\tan{\theta}}\frac{\sec^2(\theta)}{\sqrt{\sec^2(\theta)}^5}\,d\theta\\ &=&e^h\int_{\theta_0}^{\frac{\pi}{2}}e^{\tan{\theta}}\frac{\sec^2(\theta)}{\sec^5(\theta)}\,d\theta\\ &=&e^h\int_{\theta_0}^{\frac{\pi}{2}}e^{\tan{\theta}}\frac{1}{\sec^3(\theta)}\,d\theta\\ &=&e^h\int_{\theta_0}^{\frac{\pi}{2}}e^{\tan{\theta}}\cos^3(\theta)\,d\theta\\ \end{array}$$

One posible way to continue this is $\cos^3(\theta)=\cos(\theta)\cos^2(\theta)=\cos(\theta)(1-\sin(\theta))$, so

$$\begin{array}{rcl} I&=&\underbrace{e^h\int_{\theta_0}^{\frac{\pi}{2}}e^{\tan(\theta)}\cos(\theta)\,d\theta}_{I_1}+\underbrace{e^h\int_{\theta_0}^{\frac{\pi}{2}}e^{\tan(\theta)}\sin^2(\theta)\cos(\theta)\,d\theta}_{I_2} \end{array}$$

  1. For $I_1$, you could do it by parts (haven't done it myself yet)

  2. For $I_2$, you could do it by parts, plainly or trying to use first the substitution $x=\sin(\theta)$, so $dx=\cos(\theta)d\theta$ and $\tan(\theta)=\frac{x}{\sqrt{1-x^2}}$ (haven't tried it either).

Good luck!

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    $\begingroup$ $I = \int_0^\infty \cdots du = \int_{\color{red}{h}}^{\color{red}{\infty}} \cdots dt$. $\endgroup$ – achille hui Aug 14 '14 at 19:32
  • $\begingroup$ tottaly right, sorry $\endgroup$ – cjferes Aug 14 '14 at 19:37

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