<s>[INST] To simplify $f_a(x)= \int_{-a}^{+a} e^ {-\frac{x}{t^2-a^2}}\;dt$ 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.
 [/INST] 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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