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Let $x\geq 0$, then $$\int_0^{\frac{\pi}{2}} e^{-x\tan t+\alpha t} \;dt = U_{\alpha} (x) $$

$$-\int_0^{\frac{\pi}{2}} \tan t \ e^{-x\tan t+\alpha t} \;dt = \frac{d (U_{\alpha} (x) )}{dx} $$

$$\int_0^{\frac{\pi}{2}} \tan^2 t \ e^{-x\tan t+\alpha t} \;dt = \frac{d^2(U_{\alpha} (x) )}{dx^2} $$

$$\int_0^{\frac{\pi}{2}} [-x(1+\tan^2 t)+\alpha] \ e^{-x\tan t+\alpha t} \;dt = -x\frac{d^2(U_{\alpha} (x) )}{dx^2}+ (\alpha-x) U_{\alpha} (x)$$

$$ \ e^{-x\tan t+\alpha t} |_0^{\frac{\pi}{2}} = -x\frac{d^2(U_{\alpha} (x) )}{dx^2}+ (\alpha-x) U_{\alpha} (x)$$

$$ -x\frac{d^2(U_{\alpha} (x) )}{dx^2}+ (\alpha-x) U_{\alpha} (x)=-1$$

$$ x\frac{d^2(U_{\alpha} (x) )}{dx^2}+ (x-\alpha) U_{\alpha} (x)=1$$

Boundry conditions: for x=0

$$\int_0^{\frac{\pi}{2}} e^{\alpha t} \;dt = U_{\alpha} (0) $$

$$\int_0^{\frac{\pi}{2}} e^{\alpha t} \;dt = U_{\alpha} (0)= \frac{e^{\frac{\pi}{2}\alpha}-1}{\alpha}$$

for $x=\infty $ ,$U_{\alpha} (\infty)=0$

Could you please help me to find the closed form of $U_{\alpha} (x)$?

Series and special function expressions all are welcome.

Thanks a lot for answers

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I think this relates to confluent hypergeometric function (en.wikipedia.org/wiki/…). –  doraemonpaul Jun 28 '12 at 11:34
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2 Answers

up vote 4 down vote accepted

I really like this question and am curious to know the context in which such a function arises.

Denote $\delta_x = x \frac{\mathrm{d}}{\mathrm{d} x}$, and let's rewrite your differential equation with it: $\delta_x y(x) = x y^\prime(x)$ and $\delta_x^2 y(x) = x^2 y^{\prime\prime}(x) + x y^\prime(x)$, giving $$ \delta_x \left( \delta_x - 1\right) y(x) + x(x-\alpha) y(x) = x $$ Apply the annihilator of the right handside $\delta_x - 1$ to both sides yields: $$ \left( \delta_x \left( \delta_x - 1 \right)^2 + x^2 \left(\delta_x + 1\right) - \alpha x \delta_x \right) y(x) = 0 \tag{1} $$ The differential equation is of a hypergeometric type only if $\alpha = 0$, in which case the solution can be read off in terms of Meijer's G-functions: $$ y_{\alpha=0}(x) = c_1 \cdot \cos(x) + c_2 \cdot \sin(x) + c_3 \cdot G_{1,3}^{3,1}\left(\frac{x^2}{4}\left| \begin{array}{l} \frac{1}{2} \\ \frac{1}{2},\frac{1}{2},0 \\ \end{array}\right. \right) $$ Because the integral tends to zero for large $x$, $c_1=c_2=0$. The integral, for $\alpha=0$ succumbs to the Mellin convolution technique: $$ \int_0^{\pi/2} \mathrm{e}^{-x \tan(t)}\mathrm{d} t = \int_0^\infty \mathrm{e}^{-u x} \frac{1}{1+u^2}\mathrm{d} u = \int_0^\infty f_1(x u) f_2\left(\frac{1}{u}\right) \frac{\mathrm{d} u}{u} $$ where $f_1(u) = \mathrm{e}^{-u} = G_{0,1}^{1,0}\left(u\left| \begin{array}{c} 0 \\ \end{array} \right.\right)$ and $f_2(u) = \frac{u}{1+u^2} = G_{1,1}^{1,1}\left(u^2\left| \begin{array}{c} \frac{1}{2} \\ \frac{1}{2} \\ \end{array} \right. \right)$, thus getting: $$ \int_0^{\pi/2} \mathrm{e}^{-x \tan(t)}\mathrm{d} t = \frac{1}{2 \sqrt{\pi}} G_{1,3}^{3,1}\left(\frac{x^2}{4}\left| \begin{array}{l} \frac{1}{2} \\ \frac{1}{2},\frac{1}{2},0 \\ \end{array}\right. \right) = \mathrm{Ci}(x) \sin (x)+\left(\frac{\pi }{2}-\mathrm{Si}(x)\right) \cos (x) $$ where $\mathrm{Ci}(x) = - \int_{x}^\infty \frac{\cos(t)}{t} \mathrm{d} t$ and $\mathrm{Si}(x) = \int_0^x \frac{\sin(t)}{t} \mathrm{d} t$, i.e. $c_3 = \frac{1}{2 \sqrt{\pi}}$.

Notice that large $x$ asymptotic of the $\alpha=0$ solution is $y_{\alpha=0}(x) = \frac{1}{x} - \frac{2}{x^3} + \mathcal{o}\left( x^{-3} \right)$, i.e. it decreases algebraically, rather than exponentially.

Now return to equation $(1)$. Since the solution we seek decays exponentially for large $x$, its Mellin transform converges for $\Re(s)>0$: $$ \hat{y}_\alpha(s) = \int_0^\infty x^{s-1} y_\alpha(x) \mathrm{d} x $$ The differential equation $(1)$ for $y_\alpha(x)$ translates into a recurrence equation for $\hat{y}_\alpha(s)$: $$ (s+1) \left( s (s+1) \hat{y}_\alpha(s) + \hat{y}_\alpha(s) - \alpha \, \hat{y}_\alpha(s+1) \right) = 0 \tag{2} $$ Rather than solve such an equation directly, it is easier to find $\hat{y}_\alpha(s)$ directly: $$ \begin{eqnarray} \hat{y}_\alpha(s) &=& \int_0^\infty x^{s-1} y_\alpha(x) \mathrm{d} x = \int_0^{\pi/2} \mathrm{e}^{\alpha t} \int_0^{\infty} x^{s-1} \mathrm{e}^{-x \tan(t)} \mathrm{d} x \mathrm{d} t = \Gamma(s) \int_0^{\pi/2} \mathrm{e}^{\alpha t} \tan^{-s}(t) \mathrm{d} t \\ &=& \Gamma(s) \int_0^\infty \mathrm{e}^{\alpha \arctan(u)} \frac{u^{-s}}{1+u^2} \mathrm{d} u = \Gamma(s) \int_0^\infty u^{-s} \left( 1 + i u \right)^{i \frac{\alpha}{2} -1} \left( 1 - i u \right)^{-i \frac{\alpha}{2} -1} \mathrm{d} u \\ &=& \frac{\Gamma \left(1-\frac{i \alpha }{2}\right) e^{\frac{1}{2} \pi (\alpha -i s)} \Gamma (s) \Gamma (s+1) }{\alpha \Gamma \left(s-\frac{i \alpha }{2}+1\right)} {}_2F_1\left(s,-\frac{i \alpha }{2};s-\frac{i \alpha }{2}+1;-1\right) \\ &\phantom{=}&-\frac{\pi e^{-\frac{1}{2} i \pi s} \Gamma \left(1-\frac{i \alpha }{2}\right) }{\alpha \Gamma \left(-s-\frac{i \alpha }{2}+1\right) \sin(\pi s)} {}_2F_1\left(-s,-\frac{i \alpha }{2};-s-\frac{i \alpha }{2}+1;-1\right) \end{eqnarray} \tag(3) $$ where $0 < \Re(s) < 1$ is needed for convergence, in agreement with the algebraic tail of $\alpha=0$ case. Thus the solution admits Mellin-Barnes integral representation: $$ y_\alpha(x) = \frac{1}{2 \pi i}\int_{\gamma - i \infty }^{\gamma+i \infty} \hat{y}_\alpha(s) x^{-s} \mathrm{d} s $$ where $\gamma$ is arbitrary real, such that $0<\gamma<1$.

It is not hard to verify that equation $(3)$ verifies the recurrence relation $(2)$.


Series at the origin (added)

The differential equation determines the series expansion for the fundamental system, but the Mellin-Barnes representation allows you to find the series as a sum of residue. The sum of residues at $s=0$ and $s=1$ gives the following series: $$y_\alpha(z) = \frac{1}{\alpha} \left(\mathrm{e}^{\frac{\pi}{2} \alpha}-1 \right) + \mathrm{e}^{\frac{\pi}{2} \alpha} z \log(z) + \frac{z}{2\alpha} \left( -\frac{4}{\alpha+2i} {}_2F_1(1,-i \frac{\alpha}{2} ; 2-i \frac{\alpha}{2}; -1 ) + \mathrm{e}^{\frac{\pi \alpha}{2}} \left(2 i + ( 4 \gamma + i \pi -1 ) \alpha + 2 \alpha \psi\left(- i \frac{\alpha}{2} \right) \right) \right) + \mathcal{o}(z)$$

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Series at $\infty$ (added)
Having the Mellin-Barnes representation handy allows to also determine the series expansion for large $x$. One would have $y_\alpha(x) = -\sum_{k=1}^{n} \operatorname{Res}_{s=k} \hat{y}_\alpha(s) x^{-s} + \mathcal{o}\left(x^{-n}\right)$. Computing residues gives: $$ y_\alpha(x) = \frac{1}{x} + \frac{\alpha}{x^2} + \frac{\alpha^2-2}{x^3} + \frac{\alpha^3-8 \alpha}{x^4} + \mathcal{o}\left(x^{-4}\right) $$

The expansion can be easily verified to agree with the differential equation.

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Thank you very much for very detailed steps. I got many nice point. Now I wonder how can $u_{\alpha}(x)$ be expressed as series? For your information: There is no any context at least I do not know. I would like to know as well if any context that includes that function. Maybe someone else can refer. The question is my own interest and searching result. I am interested in second order differential equations and looking for different ways to find general solution of those. Thanks a lot for your answer again . –  Mathlover Jun 28 '12 at 4:59
    
I think this relates to confluent hypergeometric function (en.wikipedia.org/wiki/…). –  doraemonpaul Jun 28 '12 at 11:37
    
@Mathlover I have added the series expansion for $y_\alpha(x)$ around $x=0$. Please see the updated post. –  Sasha Jun 28 '12 at 13:16
    
@Sasha Thanks a lot for your help. very much appreciated –  Mathlover Jun 28 '12 at 13:23
    
@doraemonpaul It depends on what you mean by this. The confluent hypergeometric function satisfies the differential equation of ${}_1F_1$, i.e. $y(x) = U(a,b,x)$ satisfies $\delta_x( \delta_x + b -1 ) y(x) = x (\delta_x + a)$, which is $x y^{\prime\prime}(x) + (b-x) y^\prime(x) -a y(x) =0$, which is different from the one for $U_\alpha(x)$. –  Sasha Jun 28 '12 at 13:25
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For $\alpha=0$, Maple says this: $$ \int_{0}^{\frac{\pi}{2}} \operatorname{e} ^{-x \operatorname{tan} t} d t = \frac{i \operatorname{e} ^{i x}}{2} \mathrm{Ei}_1 (i x) - \frac{i \operatorname{e} ^{-i x}}{2} \mathrm{Ei}_1 (-ix) $$ in terms of the exponential integral $$ \mathrm{Ei}_a(z) = \int_1^\infty \operatorname{e}^{-sz} s^{-a}\,ds $$

For the DE, Maple has an answer involving integrals and the Whittaker M and W functions.

MAPLE OUTPUT

_C1 and _C2 are two arbitrary constants. It looks much worse than the original way of writing this!

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Could you please help to find the solution of differential equation too? Thanks a lot for answers –  Mathlover Jun 27 '12 at 14:38
    
Wawww. This output looks very complex. What about hypergeometric function expressions for that diff equation. After some transforms is it possible to get more simple way. thanks for advice –  Mathlover Jun 27 '12 at 14:49
    
Wolfram Alpha gave long solution with hyper-geometric functions too .wolframalpha.com/input/?i=x+u%27%27+%2B%28x-a%29u%3D1 . looks very easy differential equation but result is not. I need to find different approach to find closed form of U. Thanks for all kind of advice. –  Mathlover Jun 27 '12 at 15:26
    
Presumably the first two terms give the solution to the homogeneous equation with ${}=0$, and the integral terms are the particular solution for the ${}=1$ equation from the method of variation of parameters. –  GEdgar Jun 27 '12 at 16:18
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