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$$f(x)=\int_0^{+\infty} e^{-(t+\frac{1}{t})x}\;dt$$

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

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wolfram alpha expresses $f$ with $K_1$, the modified Bessel function of the second kind... –  Dirk Feb 23 '12 at 9:28
    
Could you please show how to get the result without wolfram alpha? Thanks –  Mathlover Feb 23 '12 at 10:01
    
With the substitution $t = e^s$ this becomes $$ f(x) = \int_{-\infty}^{\infty} e^{-(e^s+e^{-s})x}e^s ds = \frac{1}{2} \int_{-\infty}^{\infty} e^{-(e^s+e^{-s})x}(e^s+e^{-s})ds = -\frac{1}{2} f'(x). $$ So $f(x) = \alpha e^{-2x}$ for some constant $\alpha$. Not sure how to find $\alpha$. –  WimC Feb 23 '12 at 11:45

1 Answer 1

up vote 4 down vote accepted

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 $f(x)=2K_1(2x)$.

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yes, this is typo –  Norbert Feb 23 '12 at 12:23
    
Thanks a lot for answer. Great –  Mathlover Feb 23 '12 at 12:31

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