INTEGRAL xln(z^2+x^2)/e^{2 \pi x}-1\,dx

In this post we will compute the following three integrals:


\int_0^\infty\frac{x \ln(1+x^2)}{e^{\pi x}-1}\,dx =-2\zeta^{\prime}(-1)+\frac{ 2 }{3}\ln 2-\frac34


\int_0^\infty\frac{x \ln(1+x^2)}{e^{2\pi x}-1}\,dx=\zeta^{\prime}(-1)-\frac{3}{4}+\frac12 \ln 2 \pi


\int_0^\infty\frac{x \ln(1+x^2)}{e^{4\pi x}-1}\,dx=\frac{\ln \pi}{4}-\frac{36}{48}+\frac{35}{48} \ln2+\frac14 \zeta^{\prime}(-1)

Consider the following three integrals


I(z)=2\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{\pi x}-1}\,dx(1)


J(z)=2\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx(2)


K(z)=\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{4\pi x}-1}\,dx(3)

Differentiating the three w.r. to z we obtain


I^{\prime}(z)=4\int_0^\infty\frac{zx }{(z^2+x^2)(e^{\pi x}-1)}\,dx(4)


J^\prime(z)=4\int_0^\infty\frac{zx }{(z^2+x^2)(e^{2\pi x}-1)}\,dx(5)


K^{\prime}(z)=\int_0^\infty\frac{2zx }{(z^2+x^2)(e^{4\pi x}-1)}\,dx(6)


Now recall Binet´s Integral representation for the Digamma function


\int_0^\infty\frac{x }{(z^2+x^2)(e^{2\pi x}-1)}\,dx=\frac{\log(z)}{2}-\frac{\psi(z)}{2}-\frac{1}{4z}(7)


Making the change of variable 2x \to x and multiplying (7) by 4z we get


I^{\prime}(z)=4\int_0^\infty\frac{zx }{(z^2+x^2)(e^{\pi x}-1)}\,dx=2z\log\left(\frac{z}{2}\right)-2z\psi\left(\frac{z}{2}\right)-2(8)


Multiplying (3) by 4z


J^\prime(z)=4\int_0^\infty\frac{zx }{(z^2+x^2)(e^{2\pi x}-1)}\,dx=2z\log(z)-2z\psi(z)-1(9)

Integrating (8) w.r. to z

  \begin{aligned} &\int_0^z\,\int_0^\infty\frac{2wx }{(w^2+x^2)(e^{\pi x}-1)}\,dx\,dw=\int_0^zw\log\left(\frac{w}{2}\right)\,dw-\int_0^z w\psi\left(\frac{w}{2}\right)\,dw-\int_0^z dw\\ &\int_0^\infty\frac{x \ln(z^2+x^2) }{(e^{\pi x}-1)}\,dx=2\int_0^\infty\frac{x \ln(x) }{(e^{\pi x}-1)}\,dx+\frac{z^2\ln z}{2}-\frac{z^2}{4}-\frac{z^2 \ln 2 }{2}-z-4\int_0^{z/2} w\psi\left(w\right)\,dw\\ &\int_0^\infty\frac{x \ln(z^2+x^2) }{(e^{\pi x}-1)}\,dx=2\int_0^\infty\frac{x \ln(x) }{(e^{\pi x}-1)}\,dx+\frac{z^2\ln z}{2}-\frac{z^2}{4}-\frac{z^2 \ln 2 }{2}-z-4\left(\frac{z}{2} \ln\Gamma\left(\frac{z}{2}\right) -\int_0^{z/2} \ln\Gamma\left(w\right)\,dw \right)\\ \end{aligned}


Letting z \to 1


\begin{aligned} \int_0^\infty\frac{x \ln(1+x^2) }{(e^{\pi x}-1)}\,dx&=2\left(\frac16+\frac{\ln 2}{6}-2 \ln A \right)-\frac{1}{4}-\frac{ \ln 2 }{2}-1-4\left(\frac{1}{2} \ln\Gamma\left(\frac{1}{2}\right) -\int_0^{1/2} \ln\Gamma\left(w\right)\,dw \right)\\ &=\frac13-4 \ln A -\frac{5}{4}-\frac{ \ln 2 }{6}-4\left(\frac{1}{4} \ln \pi -\left( \frac32\ln A+\frac{5}{24}\ln 2+\frac14 \ln \pi\right)\right)\\ &=\frac13-4 \ln A -\frac{5}{4}-\frac{ \ln 2 }{6}+6\ln A+\frac56\ln 2\\ &=2\ln A -\frac{11}{12}+\frac{ 2 }{3}\ln 2\\ &=-2\zeta^{\prime}(-1)+\frac{ 2 }{3}\ln 2-\frac34 \qquad \blacksquare\\ \end{aligned}

Where we used that


\int_0^{1/2}\ln\left(\Gamma(x)\right)dx=\frac{3}{2}\ln A+\frac{5}{24}\ln 2+\frac{1}{4}\ln \pi

Proved here.

For the second integral, integrating from 0 to z


\begin{aligned} J(z)&=\int_0^z\,\int_0^\infty\frac{2wx }{(w^2+x^2)(e^{2\pi x}-1)}\,dx\,dw\\ &=\int_0^zw\log(w)\,dw-\int_0^zw\psi(w)\,dw-\frac12\int_0^zdw\\ &\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx-2\int_0^\infty\frac{x \ln(x)}{e^{2\pi x}-1}\,dx=\frac{z^2 \ln z}{2}-\frac{z^2}{4}-z \ln \Gamma(z)+\int_0^z\ln \Gamma(w)dw-\frac{z}{2}\\ &\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx=2\int_0^\infty\frac{x \ln(x)}{e^{2\pi x}-1}\,dx+\frac{z^2 \ln z}{2}-\frac{z^2}{4}-z \ln \Gamma(z)+\int_0^z\ln \Gamma(w)dw-\frac{z}{2}\\ &\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx=\zeta^{\prime}(-1)+\frac{z^2 \ln z}{2}-\frac{z^2}{4}-z \ln \Gamma(z)+\int_0^z\ln \Gamma(w)dw-\frac{z}{2}\\ \end{aligned}


Letting z \to 1


\int_0^\infty\frac{x \ln(1+x^2)}{e^{2\pi x}-1}\,dx=\zeta^{\prime}(-1)-\frac{3}{4}+\frac12 \ln 2 \pi \qquad \blacksquare


Where we used that

\int_0^1\ln \Gamma(w)dw=\frac{\ln(2 \pi)}{2}


Proved here.

For the last integral, We Multiply both sides of (7) by 2 \mathrm{z} to get


\int_{0}^{\infty} \frac{2 z x}{\left(z^{2}+x^{2}\right)\left(e^{2 \pi x}-1\right)} d x=z \ln (z)-z \psi(z)-\frac{1}{2}

Now let z=2 s


\begin{gathered} 2 \int_{0}^{\infty} \frac{2 s x}{\left(4 s^{2}+x^{2}\right)\left(e^{2 \pi x}-1\right)} d x=2 s \ln (2 s)-2 s \psi(2 s)-\frac{1}{2} \\ \int_{0}^{\infty} \frac{s x}{\left(s^{2}+\left(\frac{x}{2}\right)^{2}\right)\left(e^{2 \pi x}-1\right)} d x=2 s \ln (2 s)-2 s \psi(2 s)-\frac{1}{2} \end{gathered}


And now let x \rightarrow 2 x to obtain


\begin{gathered} 2 \int_{0}^{\infty} \frac{2 s x}{\left(s^{2}+x^{2}\right)\left(e^{4 \pi x}-1\right)} d x=2 s \ln (2 s)-2 s \psi(2 s)-\frac{1}{2} \\ \int_{0}^{\infty} \frac{2 s x}{\left(s^{2}+x^{2}\right)\left(e^{4 \pi x}-1\right)} d x=s \ln (2 s)-s \psi(2 s)-\frac{1}{4} \end{gathered}


Integrating from 0 to z


\begin{aligned} \int_{0}^{\infty} \frac{x \ln \left(z^{2}+x^{2}\right)}{e^{4 \pi x}-1} d x-\int_{0}^{\infty} \frac{x \ln \left(x^{2}\right)}{e^{4 \pi x}-1} d x &=\ln 2 \int_{0}^{z} s d s+\int_{0}^{z} s \ln s d s-\frac{1}{4} \int_{0}^{z} d s-\int_{0}^{z} s \psi(2 s) d s \\ \int_{0}^{\infty} \frac{x \ln \left(z^{2}+x^{2}\right)}{e^{4 \pi x}-1} d x &=2 \int_{0}^{\infty} \frac{x \ln (x)}{e^{4 \pi x}-1} d x+\ln 2 \int_{0}^{z} s d s+\int_{0}^{z} s \ln s d s-\frac{1}{4} \int_{0}^{z} d s-\int_{0}^{z} s \psi(2 s) d s \\ &=\left(\frac{1}{48}-\frac{\ln 2}{48}-\frac{12 \ln A}{48}\right)+\frac{z^{2} \ln 2}{2}+\frac{z^{2} \ln z}{2}-\frac{z^{2}}{4}-\frac{z}{4}-\frac{1}{4} \int_{0}^{2 z} s \psi(s) d s \\ &=\left(\frac{1}{48}-\frac{\ln 2}{48}-\frac{12 \ln A}{48}\right)+\frac{z^{2} \ln 2}{2}+\frac{z^{2} \ln z}{2}-\frac{z^{2}}{4}-\frac{z}{4}-\frac{1}{4}\left(z \ln \Gamma(z)-\int_{0}^{2 z} \ln \Gamma(s) d s\right) \end{aligned}


Letting z \rightarrow 1


\begin{aligned} \int_{0}^{\infty} \frac{x \ln \left(1+x^{2}\right)}{e^{4 \pi x}-1} d x &=\left(\frac{1}{48}-\frac{\ln 2}{48}-\frac{12 \ln A}{48}\right)+\frac{\ln 2}{2}-\frac{1}{4}-\frac{1}{4}+\frac{1}{4}\left(\int_{0}^{2} \ln \Gamma(s) d s\right) \\ &=-\frac{23}{48}+\frac{23 \ln 2}{48}-\frac{12 \ln A}{48}+\frac{1}{4}\left(\int_{0}^{1} \ln \Gamma(s) d s+\int_{1}^{2} \ln \Gamma(s) d s\right) \\ &=-\frac{23}{48}+\frac{23 \ln 2}{48}-\frac{12 \ln A}{48}+\frac{1}{4}\left(\int_{0}^{1} \ln \Gamma(s) d s+\int_{0}^{1} \ln \Gamma(s+1) d s\right) \\ &=-\frac{23}{48}+\frac{23 \ln 2}{48}-\frac{12 \ln A}{48}+\frac{1}{4}\left(\int_{0}^{1} \ln \Gamma(s) d s+\int_{0}^{1} \ln s \Gamma(s) d s\right) \\ &=-\frac{23}{48}+\frac{23 \ln 2}{48}-\frac{12 \ln A}{48}+\frac{1}{4}\left(2 \int_{0}^{1} \ln \Gamma(s) d s+\int_{0}^{1} \ln s d s\right) \\ &=-\frac{23}{48}+\frac{23 \ln 2}{48}-\frac{12 \ln A}{48}+\frac{1}{4}(\ln 2 \pi-1) \\ &=-\frac{23}{48}+\frac{23 \ln 2}{48}-\frac{12 \ln A}{48}+\frac{1}{4}(\ln 2+\ln \pi-1) \\ &=-\frac{23}{48}+\frac{23 \ln 2}{48}-\frac{12 \ln A}{48}+\frac{1}{48}-\frac{1}{48}+\frac{1}{4}(\ln 2+\ln \pi-1) \\ &=\frac{\ln \pi}{4}-\frac{36}{48}+\frac{35}{48} \ln 2+\frac{1}{4} \zeta^{\prime}(-1) \qquad \blacksquare \end{aligned}


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