Evaluate Integral : $\int^1_0\frac{\ln(x)}{4\pi^2+\ln^2(x)}\cdot \frac{1}{1-x}dx$

Question

Evaluate Integral : $$\int^1_0\frac{\ln(x)}{4\pi^2+\ln^2(x)}\cdot \frac{1}{1-x}dx$$

I need help to evaluate Integral Is there a way to evaluate such integrals?

Answer 1

Let $I$ denote the integral. Then

\begin{align*} I &= - \int_{0}^{\infty} \frac{t}{4\pi^2 + t^2} \frac{e^{-t}}{1 - e^{-t}} \, \mathrm{d}t \tag{$x = e^{-t}$} \\ &= - \sum_{n=1}^{\infty} \int_{0}^{\infty} \frac{t}{4\pi^2 + t^2} e^{-nt} \, \mathrm{d}t \\ &= - \sum_{n=1}^{\infty} \int_{0}^{\infty} \frac{u}{(2\pi n)^2 + u^2} e^{-u} \, \mathrm{d}u \tag{$u = nt$}\\ &= - \frac{1}{2} \int_{0}^{\infty} \left( \frac{1}{2}\coth\left(\frac{u}{2}\right) - \frac{1}{u} \right) e^{-u} \, \mathrm{d}u\\ &= - \frac{1}{2} \int_{0}^{\infty} \left( \frac{1}{e^u-1} - \frac{e^{-u}}{2} - \frac{e^{-u}}{u} \right) \, \mathrm{d}u. \end{align*}

Now we regularize the integral by considering the function

$$ I(s) = - \frac{1}{2} \int_{0}^{\infty} u^s \left( \frac{1}{e^u-1} - \frac{e^{-u}}{2} - \frac{e^{-u}}{u} \right) \, \mathrm{d}u. $$

Then it is not hard to check that $I(s)$ is analytic for $\operatorname{Re}(s) > -2$ with $I = I(0)$. Moreover, for $\operatorname{Re}(s) > 0$,

\begin{align*} I(s) &= - \frac{1}{2} \int_{0}^{\infty} \left( \frac{u^s}{e^u-1} - \frac{1}{2} u^s e^{-u} - u^{s-1}e^{-u} \right) \, \mathrm{d}u \\ &= - \frac{1}{2} \left( \Gamma(s+1)\zeta(s+1) - \frac{1}{2} \Gamma(s+1) - \Gamma(s) \right). \end{align*}

Since the right-hand side is meromorphic on $\mathbb{C}$, the principle of analytic continuation tells that this equality continues to hold for all $\operatorname{Re}(s) > -2$. Therefore, letting $s \to 0$ gives

\begin{align*} I &= \frac{1}{4} - \frac{1}{2} \lim_{s \to 0} \left( \zeta(1+s) - \frac{1}{s} \right) \\ &= \frac{1}{4} - \frac{\gamma}{2}. \end{align*}

Answer 2

Starting from @Sangchul Lee's solution

$$ I= - \sum_{n=1}^{\infty} \int_{0}^{\infty} \frac{u}{(2\pi n)^2 + u^2} e^{-u} \, du $$ consider $$J=\int \frac{u}{a^2 + u^2} e^{-u} \, du=\frac 12\int \Big(\frac{e^{-u}}{u+i a}+\frac{e^{-u}}{u-ia} \Big)\,du $$ $$J=\frac 12\Big(e^{i a}\, \text{Ei}(-u-i a)+e^{-i a}\, \text{Ei}(-u+i a) \Big)$$ For $$K=\int_0^\infty \frac{u}{a^2 + u^2} e^{-u} \, du$$ using the relation between the exponential integral function and the trigonometric integral functions $$K=\frac{1}{2} (\pi -2 \text{Si}(a)) \sin(a)-\text{Ci}(a) \cos (a)$$

$$a=2n\pi \qquad \implies \qquad K=-\text{Ci}(2 n \pi )$$

Therefore

$$I= \sum_{n=1}^{\infty}\text{Ci}(2 n \pi )=\frac{1}{4}-\frac{\gamma }{2}$$

Answer 3

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

$$I=\int_0^{\infty}\frac{x}{(e^x-1)(x^2+4\pi^2)}\,dx$$ Here's my try:

Using the below idea

$$L(\sin at)=\int_0^\infty e^{−st}\sin at\,dt$$

$$\sin at = \frac{e^{iat}-e^{-iat}}{2i}$$

$$L(\sin at)=\frac{1}{2i}\left[\int_0^\infty {e^{-t(s-ia)}\,dt-\int_0^\infty e^{-t(s+ia)}}\,dt\right]$$

$$\frac{1}{2i}\left[\int_0^\infty {e^{-t(s-ia)}\,dt-\int_0^\infty e^{-t(s+ia)}}\,dt\right]=\frac{1}{2i}\left[\frac{e^{-t(s-ia)}}{ia-s}+\frac{e^{-t(s-ia)}}{s-ia}\right]_0^\infty=\frac{a}{s^2+a^2}$$

Setting $a=x,s=2\pi$,

$$I=\int_0^\infty \frac{1}{e^x-1}\int_0^\infty e^{-2\pi t}\sin(xt)\,dt\,dx\implies \int_0^\infty\int_0^\infty e^{-2\pi t}\frac{\sin(xt)}{e^x-1}\,dx\,dt$$

Using this below result; $$\color{red}{\sum_{n=1}^\infty e^{-nx}=\frac{1}{e^x-1}}$$

$$I=\int_0^\infty\int_0^\infty e^{-2\pi t}\sin(xt)\sum_{n=1}^\infty e^{-nx} \,dx\,dt\implies \int_0^\infty\sum_{n=1}^\infty e^{-2\pi t} \int_0^\infty \sin(xt) e^{-nx} \,dx\,dt$$

$$I=\int_0^\infty e^{-2\pi t} \sum_{n=1}^\infty \frac{t}{t^2+n^2}\,dt$$

Using this, we can evaluate the sum as: $$\color{red}{\sum_{n=1}^\infty \frac{1}{t^2+n^2} =\frac12\left(\frac{\pi t \coth(\pi t)}{t^2}-\frac{1}{t^2}\right)}$$

Substitute this result above with the simplification of notation by $\pi t=z$,

$$I=\frac12\int_0^\infty e^{-2z}\left({\coth(z)}-\frac{1}{z}\right)\,dz$$

We can evalute this using integration by parts by utilizing the results $(1),(2),(3);$

$$\int \coth(x)\,dx=\ln(\sinh(x))+c\tag1$$ $$\int \frac1x\,dx=\ln(x)+c\tag2$$ $$\sinh(x)=\frac{e^x-e^{-x}}{2}\tag3$$

$$I=\int_0^\infty (\ln(\sinh(x)-\ln(x))e^{-2x}\,dx=\int_0^\infty (\ln(1-e^{-2x})-\ln(2x)+x)e^{-2x}\,dx$$

Then breaking this into three integrals and it is evaluating two of the three are straight forward, for the third one we can utilize this beautiful result;

$$\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\,dt$$

Differentiating it with respect to $z$;

$$\Gamma'(z)=\int_0^\infty t^{z-1}e^{-t}\ln(t)\,dt$$

Setting $z=1$

$$\Gamma'(1)=\int_0^\infty e^{-t}\ln(t)\,dt=-\gamma$$

The above result is from here.

And all of this nicely ends on;

Note-it checks out with this

$$\boxed{I'=-I=\frac14-\frac{\gamma}{2}\approx -0.0386}$$