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 :
Can you please help me with some technics to simplify $f_a(x)$ ?
Thanks a lot for any help.