How to compute $\int_0^\infty \frac1{(x+a)(\ln^2x+4\pi^2)}dx$

Question

Is there a closed form for this integral? $$I=\int_0^\infty \frac1{(x+a)(\ln^2x+4\pi^2)}dx,~~~a>0$$

My attempt:

I know for the case of $a=1$, there is a closed form, just do the sub $x=e^t$

$$I=\int_{-\infty}^\infty \frac{e^t}{(e^t+1)(t^2+4\pi^2)}dt=\int_{-\infty}^\infty \frac{1}{t^2+4\pi^2}dt-\int_{-\infty}^\infty \frac{1}{(e^t+1)(t^2+4\pi^2)}dt=\frac12-J$$

For $J$, do the sub $t\to-t$

$$J+J=\int_{-\infty}^\infty \left(\frac{1}{e^t+1}+\frac{1}{e^{-t}+1}\right)\frac{1}{t^2+4\pi^2}dt=\int_{-\infty}^\infty \frac{1}{t^2+4\pi^2}dt=\frac12$$

So we get

$$\int_0^\infty \frac1{(x+1)(\ln^2x+4\pi^2)}dx=\frac14$$

This is the special case (when $a=1$) due to the fact

$$\frac{1}{e^t+1}+\frac{1}{e^{-t}+1}=1$$

But what if $a\neq1$, above property doesn't hold, is there a closed form for this general parameter $a$? (Wolfram can't solve it with general parameter $a$.)

Answer 1

$$I=\int_0^\infty \frac1{(x+a)(\ln^2x+4\pi^2)}dx,~~~a>0$$

Let $x=e^t$,

$$I=\int_{-\infty}^\infty \frac{e^t}{(e^t+a)(t^2+4\pi^2)}dt=\int_{-\infty}^\infty \frac{1}{t^2+4\pi^2}dt-\int_{-\infty}^\infty \frac{a}{(e^t+a)(t^2+4\pi^2)}dt=\frac12-J$$

Next we work on $J$, note that

$$\frac{1}{t^2+4\pi^2}=\frac{1}{4\pi i}\left( \frac1{t-2\pi i}-\frac1{t+2\pi i}\right)$$

then we have

$$J=\frac a{4\pi i}\int_{-\infty}^\infty \frac{1}{e^t+a}\left( \frac1{t-2\pi i}-\frac1{t+2\pi i}\right)dt$$

Define $$f(z)=\frac1{e^z+a}\cdot \frac1z$$

and choose the rectangle contour as below, which is infinitely wide horizontally,

The integral $\int f(z) dz$ vanishes on $C_2$ and $C_4$. On $C_1$, $z=x+2\pi i$

$$\int_{C_1} f(z) dz=\int_\infty^{-\infty} \frac1{e^x+a}\cdot \frac1{x+2\pi i} ~dx=-\int^\infty_{-\infty} \frac1{e^x+a}\cdot \frac1{x+2\pi i} ~dx$$

On $C_3$, $z=x-2\pi i$

$$\int_{C_3} f(z) dz=\int_{-\infty}^\infty \frac1{e^x+a}\cdot \frac1{x-2\pi i} ~dx$$

Now we are done for the contour

$$-\int^\infty_{-\infty} \frac1{e^x+a}\cdot \frac1{x+2\pi i} ~dx+0+\int_{-\infty}^\infty \frac1{e^x+a}\cdot \frac1{x-2\pi i} ~dx+0=2\pi i \sum_{k=0}^2\text{Res}(z_k)$$ Note $z_0=0, z_1=\ln a+i\pi$ and $z_2=\ln a-i\pi$ are simple poles inside the contour, hence

$$\frac{4\pi i}{a}J=2\pi i\sum_{k=0}^2\text{Res}(z_k)=2\pi i\left( \frac{1}{1+a}+\frac{i}{a\pi-ia\ln a}+\frac{-i}{a\pi+ia\ln a} \right)$$

Simplify and we get

$$J=\frac a{2+2a}-\frac{\ln a}{\pi^2+\ln^2a}$$

Finally, $\displaystyle I=\frac12-J$, therefore,

$$\boxed{~\int_0^\infty \frac1{(x+a)(\ln^2x+4\pi^2)}dx=\frac1{2+2a}+\frac{\ln a}{\pi^2+\ln^2a}~~~~~~~~a>0~~}$$

Answer 2

Here is an elementary evaluation. Let $$I=\int_0^\infty \frac1{(x+a)(\ln^2x+4\pi^2)}dx$$ $$J(t)=\int_0^\infty \frac{x^{-t}}{(x+a)(\ln^2x+4\pi^2)}dx$$ Evaluate \begin{align} J’’(t)= \int_0^\infty \frac{x^{-t}\ln^2x}{(x+a)(\ln^2x+4\pi^2)}dx= \int_0^\infty\frac{x^{-t}}{x+a}dx-4\pi^2 J(t) \end{align} to establish $$J’’(t)+4\pi^2 J(t)=\frac\pi{a^t\sin(\pi t)}=f(t)$$ which has the general solution \begin{align} J(t)=&\ \sin(2\pi t) \bigg(c_1 +\frac1{2\pi}\int_0^t\cos(2\pi s)f(s)ds\bigg)\\ &+ \cos(2\pi t) \bigg(c_2 -\frac1{2\pi}\int_0^t\sin(2\pi s)f(s)ds\bigg) \end{align} Note that $I=J(0)$ and $$J(1)=J(0)-\int_0^1\cos(\pi s)a^{-s}ds =I-\frac{(1+a)\ln a}{a(\ln^2a+\pi^2)}\tag1 $$ On the other hand \begin{align} J(1)&= \int_0^\infty \frac{1}{x(x+a)(\ln^2x+4\pi^2)}dx\\ &= \frac1a\int_0^\infty \left(\frac{1}{x}-\frac1{x+a}\right)\frac1{\ln^2x+4\pi^2}dx=\frac1{2a}-\frac1aI\tag2 \end{align} Equate (1) and (2) to obtain $$I= \frac{\ln a}{\ln^2a+\pi^2}+ \frac1{2(1+a)} $$