# Improper integral not using any special integral

I encoutered this question- $$\int_0^{\infty} \frac{\ln(y)}{y^2+ty+t^2}dy$$ I tried to solve it using properties of definite integrals but couldn't. So then i had to solve it using indefinite integrals and then take a limit, i got- $$\dfrac{\mathrm{i}\left(\left(\ln\left(-\left(\sqrt{3}\mathrm{i}+1\right)t\right)-\ln\left(2\right)\right)\ln\left(\left|2y+\left(\sqrt{3}\mathrm{i}+1\right)t\right|\right)+\left(\ln\left(2\right)-\ln\left(\left(\sqrt{3}\mathrm{i}-1\right)t\right)\right)\ln\left(\left|2y+\left(1-\sqrt{3}\mathrm{i}\right)t\right|\right)+\operatorname{Li}_2\left(\frac{\left(\sqrt{3}\mathrm{i}+1\right)y+2t}{2t}\right)-\operatorname{Li}_2\left(-\frac{\left(\sqrt{3}\mathrm{i}-1\right)y-2t}{2t}\right)\right)}{\sqrt{3}t}$$ i am quite confident that it is correct but it took a lot of time. If you can provide me an elementary solution which doesnot require any special integrals ill be gratefull

• Do you know about residues and contour integrals? Aug 12, 2020 at 4:45
• @openspace I have found something intriguing via symmetry Aug 12, 2020 at 5:04

Use the substitution $$y=tx$$:

$$\int_0^\infty \frac{\log(tx)}{t^2x^2+t^2x+t^2}\:tdx = \frac{1}{t}\int_0^\infty \frac{\log x + \log t}{x^2+x+1}\:dx \equiv \frac{I_1+I_2}{t}$$

For $$I_1$$ under the variable interchange $$x \leftrightarrow \frac{1}{x}$$

$$I_1 = \int_0^\infty \frac{-\log x}{x^2+x+1}\:dx = -I_1 \implies I_1 = 0$$

which leaves us with

$$\frac{1}{t}I_2 = \frac{\log t}{t}\int_0^\infty \frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\:dx = \frac{2\pi\sqrt{3}\log t}{9t}$$

• Beautiful! Thanks alot
– Jack
Aug 12, 2020 at 5:26
• Shouldn't it be $I_2 \frac{1}{t}$ in the last line
– Jack
Aug 12, 2020 at 5:28
• @Jack yes thank you Aug 12, 2020 at 5:28