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

Question

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

Answer 1

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

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

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

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

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)

Answer 2

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!

Answer 3

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 $x=\sinh t\big)$.

Answer 4

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.