A FAMILY OF LOG INTEGRALS

I saw this integral today in Twitter which is related to a family of integrals, and this post is dedicated to proof the following results and find the value of some particular cases of

$\int_{0}^{\infty} \frac{\ln (x)}{1+x^{n}}d x=-\frac{\pi^{2}}{n^{2}} \cot \left(\frac{\pi}{n}\right) \csc \left(\frac{\pi}{n}\right)$

$\int_{0}^{\infty} \frac{\ln^2 (x)}{1+x^{n}} d x=\frac{\pi^3}{n^3}\csc\left( \frac{\pi}{n}\right)\left(1+2 \cot^2\left( \frac{\pi}{n}\right)\right)$

$\int_0^{\infty}\frac{\ln^3(x)}{1+x^n}dx=-\frac{\pi^4}{n^4}\left(\frac{5\cot\left( \frac{\pi}{n}\right)}{\sin\left( \frac{\pi}{n}\right)}+\frac{6 \cot^3\left( \frac{\pi}{n}\right)}{\sin\left( \frac{\pi}{n}\right)} \right)$

$\int_0^{\infty}\frac{\ln^4(x)}{1+x^n}dx=\frac{\pi^5}{n^5}\left(\frac{5}{\sin\left( \frac{\pi}{n}\right)}+\frac{10 \cot\left( \frac{\pi}{n}\right)}{\sin\left( \frac{\pi}{n}\right)} + \frac{18 \cot^2\left( \frac{\pi}{n}\right)}{\sin\left( \frac{\pi}{n}\right)}+\frac{24 \cot^4\left( \frac{\pi}{n}\right)}{\sin\left( \frac{\pi}{n}\right)} \right)$

Lets start with the first integral:

$\int_{0}^{\infty} \frac{\ln (x)}{x^{n}+1}d x=-\frac{\pi^{2}}{n^{2}} \cot \left(\frac{\pi}{n}\right) \csc \left(\frac{\pi}{n}\right)$

Consider first the following substitution $x^n=w \, \Rightarrow \, dx=w^{\frac{1}{n}-1}\frac{dw}{n}$ , than:

$\int_{0}^{\infty} \frac{\ln (x)}{x^{n}+1}d x=\frac{1}{n}\int_{0}^{\infty}\frac{w^{\frac{1}{n}-1}\ln w^{1/n}}{1+w}dw$

$=\frac{1}{n^2}\int_{0}^{\infty}\frac{w^{\frac{1}{n}-1}\ln w}{1+w}dw$ (1)

Now recall the well known result proved here

$\int_{0}^{\infty} \frac{x^{a-1}}{1+x}dx=\frac{\pi}{\sin a \pi}$

$\int_{0}^{\infty} \frac{x^{-1} e^{a \ln (x)}}{1+x}dx=\frac{\pi}{\sin a \pi}$

Differentiating both sides with respect to we get

$\int_{0}^{\infty} \frac{x^{a-1} \ln (x)}{1+x}dx=\frac{-\pi^2 \cos a \pi}{\sin^2 a \pi}=- \pi^2 \cot a \pi \csc a \pi$ (2)

Comparing (1) and (2), we see that letting $a=\frac{1}{n}$ , we obtain

$\frac{1}{n^2}\int_{0}^{\infty}\frac{w^{\frac{1}{n}-1}\ln w}{1+w}dw=-\frac{\pi^2}{n^2}\cot \left(\frac{\pi}{n}\right) \csc \left(\frac{\pi}{n}\right)$

$\boxed{\int_{0}^{\infty} \frac{\ln (x)}{1+x^{n}} d x=-\frac{\pi^{2}}{n^{2}} \cot \left(\frac{\pi}{n}\right) \csc \left(\frac{\pi}{n}\right)}$ (3)

And the first result is proved.

Now, if we let n=2 in (3) we get

$\int_{0}^{\infty} \frac{\ln (x)}{x^{2}+1} d x=-\frac{\pi^{2}}{4}} \cot \left(\frac{\pi}{2}\right) \csc \left(\frac{\pi}{2}\right)$

$\boxed{\int_{0}^{\infty} \frac{\ln (x)}{x^{2}+1} d x=0}$ (4)

since $\cot \left(\frac{\pi}{2}\right)=0$ .

for n=3

$\int_{0}^{\infty} \frac{\ln (x)}{x^{3}+1} d x=-\frac{\pi^{2}}{9}} \cot \left(\frac{\pi}{3}\right) \csc \left(\frac{\pi}{3}\right)$

$\int_{0}^{\infty} \frac{\ln (x)}{x^{3}+1} d x=-\frac{\pi^{2}}{9}} \cdot \frac{1}{\sqrt{3}} \cdot \frac{2}{\sqrt{3}}$

$\boxed{\int_{0}^{\infty} \frac{\ln (x)}{x^{3}+1} d x=-\frac{2 \pi^{2}}{27}}}$ (5)

for n=4

$\int_{0}^{\infty} \frac{\ln (x)}{x^{4}+1} d x=-\frac{\pi^{2}}{16}} \cot \left(\frac{\pi}{4}\right) \csc \left(\frac{\pi}{4}\right)$

$\boxed{\int_{0}^{\infty} \frac{\ln (x)}{x^{4}+1} d x=-\frac{\pi^{2}\sqrt{2}}{16}}}$ (6)

Now consider the second integral, namely:

$\int_{0}^{\infty} \frac{\ln^2 (x)}{1+x^{n}} d x=\frac{\pi^3}{n^3}\csc\left( \frac{\pi}{n}\right)\left(1+2 \cot^2\left( \frac{\pi}{n}\right)\right)$

let $x^n=w \, \Rightarrow \, dx=w^{\frac{1}{n}-1}\frac{dw}{n}$

$\int_{0}^{\infty} \frac{\ln^2 (x)}{x^{n}+1} d x=\frac{1}{n}\int_{0}^{\infty} \frac{w^{\frac{1}{n}-1}\ln^2 (w^{1/n})}{1+w}dw$

$=\frac{1}{n^3}\int_{0}^{\infty} \frac{w^{\frac{1}{n}-1}\ln^2 (w)}{1+w}dw$ (7)

on the other hand from (2)

$\int_{0}^{\infty} \frac{x^{a-1} \ln (x)}{1+x}dx=\frac{-\pi^2 \cos a \pi}{\sin^2 a \pi}$

differentiating the above equation with respect to

$\int_{0}^{\infty} \frac{x^{a-1} \ln^2 (x)}{1+x}dx=\pi^3\left(\frac{\sin^3(a \pi)+ 2\cos^2 (a \pi)\sin(a \pi)}{\sin^4 a \pi}\right)$

$\int_{0}^{\infty} \frac{x^{a-1} \ln^2 (x)}{1+x}dx=\pi^3\left(\frac{1}{\sin a \pi}+2 \frac{\cot^2(a \pi)}{\sin(a \pi)}\right)$

$\int_{0}^{\infty} \frac{x^{a-1} \ln^2 (x)}{1+x}dx=\pi^3\left(\csc(a \pi)+2 \cot^2(a \pi)\csc(a \pi)\right)$

$\boxed{\int_{0}^{\infty} \frac{x^{a-1} \ln^2 (x)}{1+x}dx=\pi^3 \csc(a \pi)\left(1+2 \cot^2(a \pi)\right)}$ (8)

Comparing (7) and (8), and setting $a=\frac{1}{n}$ and the second result is established

$\boxed{\int_{0}^{\infty} \frac{\ln^2 (x)}{x^{n}+1} d x=\frac{\pi^3}{n^3}\csc\left( \frac{\pi}{n}\right)\left(1+2 \cot^2\left( \frac{\pi}{n}\right)\right)}$ (9)

for n=2

$\boxed{\int_{0}^{\infty} \frac{\ln^2 (x)}{x^{2}+1} d x=\frac{\pi^3}{8}}$ (10)

for n=3

$\int_{0}^{\infty} \frac{\ln^2 (x)}{x^{3}+1} d x=\frac{\pi^3}{27} \cdot \frac{2}{\sqrt{3}}\left(1+ \frac{2}{3}\right)$

$\boxed{\int_{0}^{\infty} \frac{\ln^2 (x)}{x^{3}+1} d x=\frac{10 \pi^3}{81\sqrt{3}} }$ (11)

for n=4

$\int_{0}^{\infty} \frac{\ln^2 (x)}{x^{4}+1} d x=\frac{\pi^3}{64}\csc\left( \frac{\pi}{4}\right)\left(1+2 \cot^2\left( \frac{\pi}{4}\right)\right)$

$=\frac{\pi^3 \sqrt{2}}{64}\left(1+2 \right)\right)$

$\boxed{\int_{0}^{\infty} \frac{\ln^2 (x)}{x^{4}+1} d x=\frac{3 \pi^2}{32 \sqrt{2}}}$ (12)

The third integral

letting x^n=w we obtain

$\int_0^{\infty}\frac{\ln^3(x)}{1+x^n}dx=\frac{1}{n}\int_0^{\infty}\frac{w^{1/n-1}\ln^3(w^{1/n})}{1+w}dw$

$= \,\frac{1}{n^4}\int_0^{\infty}\frac{w^{1/n-1}\ln^3(w)}{1+w}dw$

Differentiating both sides of (8) with respect to we get

$\int_{0}^{\infty} \frac{x^{a-1} \ln^3 (x)}{1+x}dx=-\pi^4\left(\frac{\cot(a \pi)}{\sin(a \pi)}+2\left(\frac{2 \sin^4(a \pi)\cos(a \pi)+3\sin^2(a \pi)\cos^3(a \pi)}{\sin^6(a \pi)} \right)\right)$

$\int_{0}^{\infty} \frac{x^{a-1} \ln^3 (x)}{1+x}dx=-\pi^4\left(\frac{\cot(a \pi)}{\sin(a \pi)}+\frac{4 \cos(a \pi)}{\sin^2(a \pi)}+\frac{6 \cos^3(a \pi)}{\sin^4(a \pi)} \right)$

$\int_{0}^{\infty} \frac{x^{a-1} \ln^3 (x)}{1+x}dx=-\pi^4\left(\frac{5\cot(a \pi)}{\sin(a \pi)}+\frac{6 \cot^3(a \pi)}{\sin(a \pi)} \right)$ (13)

and we obtain

$\boxed{\int_0^{\infty}\frac{\ln^3(x)}{1+x^n}dx=-\frac{\pi^4}{n^4}\left(\frac{5\cot\left( \frac{\pi}{n}\right)}{\sin\left( \frac{\pi}{n}\right)}+\frac{6 \cot^3\left( \frac{\pi}{n}\right)}{\sin\left( \frac{\pi}{n}\right)} \right)}$ (14)

for n=2

$\boxed{\int_0^{\infty}\frac{\ln^3(x)}{1+x^2}dx=0}$ (15)

for n=3

$\boxed{\int_0^{\infty}\frac{\ln^3(x)}{1+x^3}dx=-\frac{14 \pi^4}{243}}$ (16)

for n=4

$\boxed{\int_0^{\infty}\frac{\ln^3(x)}{1+x^4}dx=-\frac{11 \pi^4}{128 \sqrt{2}}}$ (17)

For the last integral

let x^n=w

$\int_0^{\infty}\frac{\ln^4(x)}{1+x^n}dx= \,\frac{1}{n}\int_0^{\infty}\frac{w^{1/n-1}\ln^4(w^{1/n})}{1+w}dw$

$= \,\frac{1}{n^5}\int_0^{\infty}\frac{w^{1/n-1}\ln^4(w)}{1+w}dw$ (18)

Differentiating both sides of (13) with respect to a we get

$\int_{0}^{\infty} \frac{x^{a-1} \ln^4 (x)}{1+x}dx=-\pi^4\left(5\left(\frac{-\pi \sin^3(a \pi)-2\pi \sin(a \pi) \cos^2(a \pi)}{\sin^4(a \pi)} \right)+6\left(\frac{3 \sin^5(a \pi) \cos^2(a \pi)+4 \sin^3(a \pi) \cos^4(a \pi)}{\sin^8(a \pi)} \right) \right)$

$\int_{0}^{\infty} \frac{x^{a-1} \ln^4 (x)}{1+x}dx=\pi^5\left(5\left(\frac{1}{\sin(a \pi)}+\frac{2 \cos^2(a \pi)}{\sin^3(a \pi)} \right)+6\left( \frac{3 \cos^2(a \pi)}{\sin^3(a \pi)}+\frac{4 \cos^4(a \pi)}{\sin^5(a \pi)}\right) \right)$

$\int_{0}^{\infty} \frac{x^{a-1} \ln^4 (x)}{1+x}dx=\pi^5\left(\frac{5}{\sin(a \pi)}+\frac{10 \cot(a \pi)}{\sin(a \pi)} + \frac{18 \cot^2(a \pi)}{\sin(a \pi)}+\frac{24 \cot^4(a \pi)}{\sin(a \pi)} \right)$ (19)

Comparing (18) and (19) we conclude that

$\boxed{\int_0^{\infty}\frac{\ln^4(x)}{1+x^n}dx=\frac{\pi^5}{n^5}\left(\frac{5}{\sin\left( \frac{\pi}{n}\right)}+\frac{10 \cot\left( \frac{\pi}{n}\right)}{\sin\left( \frac{\pi}{n}\right)} + \frac{18 \cot^2\left( \frac{\pi}{n}\right)}{\sin\left( \frac{\pi}{n}\right)}+\frac{24 \cot^4\left( \frac{\pi}{n}\right)}{\sin\left( \frac{\pi}{n}\right)} \right)}$ (20)

If we let n=2 we obtain

$\boxed{\int_0^{\infty}\frac{\ln^4(x)}{1+x^2}dx=\frac{5 \pi^5}{32}}$ (21)

If we let n=3 we get

$\boxed{\int_0^{\infty}\frac{\ln^4(x)}{1+x^3}dx=\frac{ 2\pi^5}{243\sqrt{3}}\left(11+\frac{10}{\sqrt{3}}+\frac{8}{3} \right)}$ (22)

If we let n=4 we obtain

$\boxed{\int_0^{\infty}\frac{\ln^4(x)}{1+x^4}dx=\frac{57\pi^5}{512 \sqrt{2}}}$ (23)

Ricardo Albahari

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A FAMILY OF LOG INTEGRALS

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