TITLE: $f(x)=\int_{0}^{+\infty} e^{-(t+\frac{1}{t})x}dt$ how to find $f(x)$?
QUESTION [9 upvotes]: $$f(x)=\int_0^{+\infty}  e^{-(t+\frac{1}{t})x}\;dt$$
if while  $ x>0 $ ,  $ f(x) $  has values
I noticed some interesting relations for $f(x)$ as shown below:
$$
\begin{align}
t & =\frac{1}{z} \\
f(x) & =\int_0^{+\infty} \frac{1}{z^2} e^{-(z+\frac{1}{z})x}dz=\int_0^{+\infty} \frac{1}{t^{2}} e^{-(t+\frac{1}{t})x} \; dt \\
f'(x) & =-\int_{0}^{+\infty} (t+\frac{1}{t}) e^{-(t+\frac{1}{t})x}\;dt \\
f''(x) & = \int_0^{+\infty} (t^2+2+\frac{1}{t^2}) e^{-(t+\frac{1}{t})x}\;dt=\int_0^{+\infty} (t^{2}+\frac{1}{t^2}) e^{-(t+\frac{1}{t})x}dt+2\int_0^{+\infty}  e^{-(t+\frac{1}{t})x}\;dt \\
f''(x) & =\int_0^{+\infty} (t^2+\frac{1}{t^2}) e^{-(t+\frac{1}{t})x}\;dt+2f(x) \\
f''(x)-2f(x) & =\int_0^{+\infty} (t^2+\frac{1}{t^2}) e^{-(t+\frac{1}{t})x} \; dt=\int_0^{+\infty} t^2 e^{-(t+\frac{1}{t})x}\;dt+\int_0^{+\infty} \frac{1}{t^2} e^{-(t+\frac{1}{t})x}\;dt \\
f''(x)-2f(x) & =\int_0^{+\infty} t^2 e^{-(t+\frac{1}{t})x}dt+f(x) \\
f''(x)-3f(x) & =\int_{0}^{+\infty} t^{2} e^{-(t+\frac{1}{t})x}dt
\end{align}
$$
and also another relation
$$f(x)=\int_0^{+\infty}  e^{-(t+\frac{1}{t})x}dt=\int_0^1 e^{-(t+\frac{1}{t})x}\;dt+\int_1^{+\infty}  e^{-(t+\frac{1}{t})x}\;dt$$
$$=\int_0^1 e^{-(t+\frac{1}{t})x}\;dt+\int_0^1 \frac{1}{z^2}e^{-(z+\frac{1}{z})x}\;dz=\int_0^1(1+\frac{1}{z^2})e^{-(z+\frac{1}{z})x}\;dz$$
I can write many relation that related to that function However I havent expressed it yet as a known function relation.
How can I find $f(x)$?
Thanks a lot for answers

REPLY [5 votes]: Note that
$$
f(x)=\int\limits_{(0,+\infty)}e^{-x(t+1/t)}dt=
\{t=e^u\}=\int\limits_{(-\infty,+\infty)}e^{-x(e^u+e^{-u})}e^u du=
$$
$$
\int\limits_{(-\infty,0)}e^{-x(e^u+e^{-u})}e^u du+\int\limits_{(0,+\infty)}e^{-x(e^u+e^{-u})}e^u du
$$
Now consider the first integral
$$
\int\limits_{(-\infty,0)}e^{-x(e^u+e^{-u})}e^u du=\{v=-u\}=
\int\limits_{(+\infty,0)}e^{-x(e^{-v}+e^v)}e^{-v} (-dv)=
$$
$$
\int\limits_{(0,+\infty)}e^{-x(e^v+e^{-v})}e^{-v} dv=
\int\limits_{(0,+\infty)}e^{-x(e^{u}+e^{-u})}e^{-u} du=
$$
So
$$
f(x)=\int\limits_{(-\infty,0)}e^{-x(e^u+e^{-u})}e^u du+\int\limits_{(0,+\infty)}e^{-x(e^u+e^{-u})}e^u du=
$$
$$
\int\limits_{(0,+\infty)}e^{-x(e^u+e^{-u})}(e^u+e^{-u}) du=
2\int\limits_{(0,+\infty)}e^{-2x\cosh(u)}\cosh(u) du
$$
From the formula given on wikipedia
$$
K_\alpha(x)=\int\limits_{(0,+\infty)}e^{-x\cosh(t)}\cosh(\alpha t)dt
$$
hence $f(x)=2K_1(2x)$.