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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

Harry Peter
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SSC Napoli
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  • 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. – Aaron Aug 25 '14 at 16:53
  • Could you post the first steps in an answer so ti make clear What you mean..? – SSC Napoli Aug 25 '14 at 16:57
  • 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. – Aaron Aug 25 '14 at 17:49

4 Answers4

2

$\int_0^\infty e^{-u}\dfrac{1}{\left(\sqrt{1+(h+u)^2}\right)^5}du$

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

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

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

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)

Harry Peter
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  • 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 $K_{-1}$ – SSC Napoli Sep 05 '14 at 18:52
  • @user3810266 Note that the incomplete bessel K function defined in http://artax.karlin.mff.cuni.cz/r-help/library/DistributionUtils/html/incompleteBesselK.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. – Harry Peter Sep 09 '14 at 12:15
  • 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 $K_{\nu}(x,y)$ – SSC Napoli Sep 11 '14 at 11:22
1

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 $x=\sinh t\big)$.

Lucian
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  • 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? – SSC Napoli Aug 15 '14 at 01:43
  • @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$\big)$. – Lucian Aug 15 '14 at 01:48
  • 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? – SSC Napoli Aug 15 '14 at 01:51
  • @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. – Lucian Aug 15 '14 at 02:04
  • Now it'a celar...Anyway Could you please give me a reference for this derivation..? – SSC Napoli Aug 15 '14 at 02:09
  • @user3810266: See, for instance, formula $11.5.2$ here for $z=1$ and $\nu=-2$, then express the Struve K function in terms of the Struve H and Bessel Y functions, using formula $11.2.5$ here. – Lucian Aug 15 '14 at 03:08
  • I still don't understand Two things first:why if i have $x=h+u$ that change the lower interval of the ingegrand. Second: in formula 11.5.2 that you posted there's a gamma function however in the solution that you posted here there's no gamma function... – SSC Napoli Aug 15 '14 at 13:22
  • @user3810266: If we substitute $t=u+h$, then the new limits will be $0+h=h$ and $\infty+h=\infty$. $\Gamma(1/2)=\sqrt\pi~$, and $\Gamma(1+x)=x\cdot\Gamma(x)$. – Lucian Aug 15 '14 at 14:08
  • Thanks alot again. Anyway i googled "incomplete struve function" and i found a reference: M.M. Agrest and M.S. maksimov Theory of Incomplete cylindrical Functions and their applications" could that be somewhat useful? – SSC Napoli Aug 15 '14 at 14:15
  • @user3810266: Unfortunately there don't seem to be any free versions online, so I can't really say. – Lucian Aug 15 '14 at 14:29
  • I think your claim is basing on http://dlmf.nist.gov/10.9#E5, but I think you are get trapped by the order of $\dfrac{5}{2}$ of $1+x^2$, since unfortunately the result only suitable for $Re(v)>-\dfrac{1}{2}$ . – Harry Peter Aug 28 '14 at 15:59
  • @HarryPeter: I always verify my results, both numerically and symbolically, using Maple and Mathematica. This one is no different. – Lucian Aug 28 '14 at 16:12
  • Please give the detail proof of this, otherwise we are very hard to believe. – Harry Peter Sep 02 '14 at 14:01
  • @user3810266 The definition of incomplete bessel K function like http://artax.karlin.mff.cuni.cz/r-help/library/DistributionUtils/html/incompleteBesselK.html can combine the scope of incomplete bessel-type function and incomplete struve-type function. – Harry Peter Sep 02 '14 at 14:05
  • @HarryPeter: The reason for which the formula you linked at is only true for $\nu>-\dfrac12$ is because the integral containing the sine function does not converge for smaller values. But this is simply not the case here, since the integral from this post converges for all real values of the denominator's exponent. Which is also why the formula that I linked at imposes no condition on $\nu$. – Lucian Sep 02 '14 at 14:50
  • thanks alot for all the comments, by looking at the link posted by Lucian it seems that there's no restriction for the exponent. However i'm very interested in the result obtained by Harry Peter it looks like at the end the incomplete Bessel function comes out as predicted by Lucian...can we verify the solution somehow? – SSC Napoli Sep 02 '14 at 17:09
  • But even you claim is basing on http://dlmf.nist.gov/11.5#E2, do you really make sure the result is also suitable for $\text{Re}(\nu)<-\dfrac{1}{2}$ because of involving the gamma function of negative number? While the result of http://math.stackexchange.com/questions/554314 is only suitable for $\text{Re}(\nu)>-\dfrac{1}{2}$ , I don't think http://dlmf.nist.gov/11.5#E2 is really suitable for $\text{Re}(\nu)<-\dfrac{1}{2}$ . – Harry Peter Sep 05 '14 at 15:51
  • Actually according to link the gamma function can have a negative argument as long as it's not a negative integer. In this case the argument of the gamma function would be $-2+\frac{1}{2}$ that is a valid argument. Anyway can we verify the solution that you posted somehow? i could try to fix $h$ equal to a certain value and perform a numerical integration of the integral that i posted. Then we can substitute the chosen value of $h$ into your solution and compare the two results... – SSC Napoli Sep 05 '14 at 16:14
  • for h=1 your derivation seems correct, both integrals give as result: $0.0522236$ – SSC Napoli Sep 05 '14 at 18:48
  • @user3810266: As I also said to Harry, I already did all this before posting the solution. – Lucian Sep 05 '14 at 21:11
  • Lucian your answer is correct when $h=0$, however the solution posted by Harry works even if $h \neq 0$ as i checked numerically... I also expect that Harry's solution to collapse to yours when $h=0$ – SSC Napoli Sep 05 '14 at 21:19
  • @user3810266: I believe you are confusing things. The integral expression of the $\Gamma$ function diverges for ALL negative values of the argument. As does also the sine integral in the link provided by Harry when $\nu<-\dfrac12$. But this is not the same as saying that the function itself cannot be extended analytically to cover values which its original expression was unable to. However, the existence of such an analytical extension does NOT mean that the afore-mentioned expression converges. Indeed, it doesn't. E.g., the p-series diverges when $p<1$, but the $\zeta$ function does not. – Lucian Sep 05 '14 at 22:36
0

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.

Aaron
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  • a comment here, is $y=y(x)$? and once I determined $y$ how do i get the final solution to the original integral? – SSC Napoli Sep 02 '14 at 17:18
  • $y$ 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 $I(\infty)$. – Aaron Sep 02 '14 at 19:00
  • unfortunately there doesn't seem to be an analytical expression for the indefinite integral...at least according to mathematica – SSC Napoli Sep 03 '14 at 15:10
-1

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!

cjferes
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