Evaluate
$$\int_0^\infty\left(\frac{\tanh x}{x^2}-\frac{1}{xe^{2x}}\right)dx$$
I haven't been able to find references to the indefinite integral of the $\tanh$ term except for some similar forms that had solutions. see here and here.
Edit:
Following Random Variable's result we have the form
$$12\log A-\frac{4}{3}\log 2$$
Answer
Let $$I(z) =\frac{z}{a}\int_0^\infty \left(\frac{1}{2}-\frac{z}{at}+\frac{1}{e^{at/z}-1}\right) \frac{1-e^{-at}}{t^2} dt$$
Then Binet's first formula says
$$I(z) = \int_0^z \left[\ln\Gamma(x) - (z-\frac{1}{2})\ln z + z - \frac{\ln(2\pi)}{2}\right] dx $$
Letting $a=2,z=1/2$ gives
$$4I(\frac{1}{2}) =\int_0^\infty \left(\frac{1}{2}-\frac{1}{4t}+\frac{1}{e^{4t}-1}\right) \frac{1-e^{-2t}}{t^2} dt$$
and $a=4,z=1$ gives
$$4I(1) =\int_0^\infty \left(\frac{1}{2}-\frac{1}{4t}+\frac{1}{e^{4t}-1}\right) \frac{1-e^{-4t}}{t^2} dt$$
Some algebraic manipulation yields
$$4I(1)-4I(\frac{1}{2}) = \underbrace{\int_0^\infty \left[\frac{1}{2t^2}-\frac{e^{-2t}}{2t} - (\frac{1}{2}-\frac{1}{4t})\left(\frac{e^{-2t}-e^{-4t}}{t^2}\right)\right] dt}_{J} - \frac{I}{2}$$
With $I$ your desired integral. Surprisingly, $J$ has elementary primitive (even it were not, we still have systematic way to crash it), with value $5/8$.
Hence it remains to evalute $$\int_0^1 \ln \Gamma(x) dx \quad \quad \int_0^{1/2} \ln \Gamma(x) dx$$
The former is just $\ln(2\pi)/2$, for the latter, we can use the integral representation of Barnes G function:
$$\int_0^z \ln \Gamma(x) dx = \frac{z(1-z)}{2}+\frac{z}{2}\ln(2\pi) + z \ln\Gamma(z) - \log G(1+z)$$
and the special value $$\ln G(\frac{3}{2}) = -\frac{3}{2}\ln A + \frac{\ln \pi}{4}+\frac{1}{8}+\frac{\ln 2}{24}$$
with $A$ being the Glaisher-Kinkelin constant.
Alternatively, use Fourier expansion of $\ln \Gamma(x)$, integrate termwise, and remember the relation between $A$ and $\zeta'(2)$ also gives the value of the integral $$\int_0^{1/2} \ln\Gamma(x)dx = \frac{3}{2}\ln A + \frac{5}{24}\ln 2 + \frac{\ln \pi}{4}$$
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