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Let $x\leq0$, then $$ f_a(x)= \int_{-a}^{+a} e^ {-\frac{x}{t^2-a^2}}\;dt$$

$$ f'(x)= -\int_{-a}^{+a} \frac{1}{t^2-a^2} e^ {-\frac{x}{t^2-a^2}}dt$$

$$ f'(x)= -\int_{-a}^{+a} \frac{t^2-(t^2-a^2)}{a^2(t^2-a^2)} e^ {-\frac{x}{t^2-a^2}}dt$$

$$ f'(x)= -\int_{-a}^{+a} \frac{t^2}{a^2(t^2-a^2)} e^ {-\frac{x}{t^2-a^2}}dt + \int_{-a}^{+a} \frac{1}{a^2} e^ {-\frac{x}{t^2-a^2}}dt$$

$$ f'(x)= -\int_{-a}^{+a} \frac{t^2}{a^2(t^2-a^2)} e^ {-\frac{x}{t^2-a^2}}dt + \frac{f(x)}{a^2} $$

$$ f''(x)= \int_{-a}^{+a} \frac{t^2}{a^2(t^2-a^2)^2} e^ {-\frac{x}{t^2-a^2}}dt + \frac{f'(x)}{a^2} $$

$$ f''(x)= \frac{1 }{2xa^2}\int_{-a}^{+a} t\frac{2xt }{(t^2-a^2)^2} e^ {-\frac{x}{t^2-a^2}}dt + \frac{f'(x)}{a^2} $$

$$ f''(x)= \frac{1 }{2xa^2}(te^ {-\frac{x}{t^2-a^2}}|{_{-a}^{+a}})-\frac{1 }{2xa^2}\int_{-a}^{+a} e^ {-\frac{x}{t^2-a^2}}dt + \frac{f'(x)}{a^2} $$

$$ f''(x)= -\frac{f(x) }{2xa^2} + \frac{f'(x)}{a^2} $$

$$ a^2f''(x) - f'(x)+\frac{f(x) }{2x}=0 $$

wolframalpha gave me that long result :

enter image description here

Can you please help me with some technics to simplify $f_a(x)$ ?

Thanks a lot for any help.

  • 0
    It seems $f_a(x)$ is infinite when $a\ne0$ and $x\gt0$.2012-06-21
  • 0
    Yes you are right typo should be x<=0 . Thanks a lot.2012-06-21

1 Answers 1

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How about $$ f_a(x) = \frac{\operatorname{e} ^{x/(2a^2)} x}{a} \biggl(\mathrm K_0 \biggl(\frac{-x}{2 a^{2}}\biggr) - \mathrm K_1 \biggl(\frac{-x}{2 a^{2}}\biggr)\biggr),\qquad x<0 $$ in terms of Bessel functions.

added

It should work like this: The function $f$ listed on the right-hand-side satisfies the differential equation. And the "initial conditions" $$ f(0) = 2a,\qquad f'(x) \sim \frac{-1}{a}\,\log\left(-x\right)\qquad\text{as } x \uparrow 0 $$ match $f_a$, therefore this one is equal to $f_a$.

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    Could you please give more detail how you got that result? Thanks2012-06-22