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I am trying to prove that

$$\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$$

I know how to deal with integrals involving cyclotomic polynomials and nested logarithms but I have no idea with this one.

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Nice question. (+1) –  Chris's sis Jan 23 '13 at 17:14

3 Answers 3

up vote 20 down vote accepted

Let's introduce the parameter $\alpha$, and then differentiate with respect to $\alpha$ that yields $$I(\alpha)=\int_0^1 \frac{t^\alpha-1}{(t^2+1)\ln t}dt $$ $$I'(\alpha)=\int_0^1 \frac{t^{\alpha}}{(t^2+1)}dt=\frac{1}{4} \left(-\psi_0\left(\frac{1 + \alpha}{4}\right) + \psi_0\left(\frac{3 + \alpha}{4}\right)\right) $$ Then $$I(\alpha)=\frac{1}{4} \int\left(-\psi_0\left(\frac{1 + \alpha}{4}\right) + \psi_0\left(\frac{3 + \alpha}{4}\right)\right) d\alpha= $$ $$I(\alpha)=\left(\ln \Gamma \left(\frac{3 + \alpha}{4}\right)- \ln \Gamma \left(\frac{1 + \alpha}{4}\right)\right)+C\tag1$$ If letting $\alpha=2$, then $$I(2)=\ln \left(\frac{\Gamma \left(\frac{5}{4}\right)}{\Gamma \left(\frac{3}{4}\right)}\right)+C$$ On the other hand, by letting $\alpha=0$ in $(1)$ we get $$C=\ln \left(\frac{\Gamma \left(\frac{1}{4}\right)}{\Gamma \left(\frac{3}{4}\right)}\right)$$ Thus $$\int_0^1 \frac{t^2-1}{(t^2+1)\ln t}dt=\ln \left(\frac{\Gamma \left(\frac{5}{4}\right)\Gamma \left(\frac{1}{4}\right)}{\Gamma^2 \left(\frac{3}{4}\right)}\right) $$

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Nice answer (+1). –  Mhenni Benghorbal Jan 23 '13 at 19:13
    
@MhenniBenghorbal: thank you! :-) –  Chris's sis Jan 23 '13 at 19:14
    
Your final answer seems to be off by a factor of 2. –  robjohn Jan 23 '13 at 21:07
    
@Chris'ssister: doh... I missed the square on the inside. Yes, your answer matches all the others :-) –  robjohn Jan 23 '13 at 21:49
    
Thank You! This was very helpful. –  Integrals and Series Jan 24 '13 at 8:13

$$ \begin{align} \int_0^1\frac{t+1}{t^2+1}\frac{t-1}{\log(t)}\,\mathrm{d}t &=\int_0^1\frac{t+1}{t^2+1}\int_0^1t^x\,\mathrm{d}x\,\mathrm{d}t\\ &=\int_0^1\int_0^1\frac{t^{x+1}+t^x}{t^2+1}\,\mathrm{d}t\,\mathrm{d}x\\ &=\int_0^1\left(\frac1{x+1}+\frac1{x+2}-\frac1{x+3}-\frac1{x+4}+\dots\right)\,\mathrm{d}x\\ &=\left(\log\left(\frac21\right)+\log\left(\frac32\right)\right) -\left(\log\left(\frac43\right)+\log\left(\frac54\right)\right)+\dots\\ &=\log\left(\frac31\right)-\log\left(\frac53\right)+\log\left(\frac75\right)-\log\left(\frac97\right)+\log\left(\frac{11}9\right)-\dots\\ &=\log\left(\frac31\cdot\frac35\cdot\frac75\cdot\frac79\cdot\frac{11}9\cdots\right)\\ &=\lim_{n\to\infty}\log\left(\frac{\Gamma\left(\frac54\right)^2}{\Gamma\left(\frac34\right)^2} \frac{\Gamma\left(\frac{4n+3}4\right)^2}{\Gamma\left(\frac{4n+5}4\right)^2}(4n+3)\right)\\ &=2\log\left(2\frac{\Gamma\left(\frac54\right)}{\Gamma\left(\frac34\right)}\right) \end{align} $$ The last equality is due to Gautschi's inequality.

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A powerful answer! I love the use of the double integrals technique that usually works great to Frullani's integrals (+1) –  Chris's sis Jan 23 '13 at 21:24
    
@Chris'ssister: I actually started using $\lim\limits_{n\to\infty}n(x^{1/n}-1)=\log(x)$, but $$\frac1n\frac{t-1}{t^{1/n}-1}=\frac1n\left(t^{(n-1)/n}+t^{(n-2)/n}+t^{(n-3)/n}+‌​\dots+1\right)$$ is the Riemann sum for $\int_0^1t^x\,\mathrm{d}t$, so I just used that. :-) –  robjohn Jan 23 '13 at 21:58
    
Nice. I was only saying that $\int_0^1 \frac{t-1}{\log(t)}\,\mathrm{d}t$ is a Frullani integral and then may be nicely tackled by double integrals. It's exactly what you did. :) –  Chris's sis Jan 23 '13 at 22:04

There have been similar integrals such as this with a factor of $1/\log{t}$ in the integrand. The way I have attacked these is to use the substitution $t=e^{-x}$; here, this produces

$$-\int_0^{\infty} dx \: \frac{e^{-x}}{x} \frac{1-e^{-2 x}}{1+e^{-2 x}} $$

$$ = -\int_0^{\infty} dx \: \frac{1}{x} (e^{-x} - e^{-3 x}) \sum_{k=0}^{\infty} (-1)^k e^{-2 k x} $$

Now, we reverse the order of the sum and integral. This is justified by Abel's theorem, although I will leave the details for another time or another person to fill in:

$$ = -\sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} dx \: \frac{1}{x} (e^{-(2 k+1) x} - e^{-(2 k+3) x}) $$

The integrals have a simple closed form, and the result is now a sum:

$$ = \sum_{k=0}^{\infty} (-1)^{k+1} \log{ \left [ \frac{(2 k+1)}{(2 k+3)} \right ] } $$

You can form a log of a Wallis-type product from this sum:

$$ = \log{\left [ \prod_{k=0}^{\infty} \frac{4 k+3}{4 k+1} \frac{4 k+3}{4 k+5} \right ]} $$

Partial products of the above product may be evaluated:

$$ \prod_{k=0}^{n} \frac{4 k+3}{4 k+1} \frac{4 k+3}{4 k+5} = \frac{9 \Gamma{\left ( \frac{5}{4} \right )} \Gamma{\left ( \frac{9}{4} \right )} \Gamma{\left ( n + \frac{7}{4} \right )}^2}{5 \Gamma{\left ( \frac{7}{4} \right )}^2 \Gamma{\left (n+ \frac{5}{4} \right )} \Gamma{\left ( n + \frac{9}{4} \right )}}$$

You may show that the above expression converges as $n \rightarrow \infty$. A little manipulation produces the stated result.

EDIT

@MikeSpivey points my attention to an equation out of Whittaker & Watson that applies here:

$$ \prod_{k=0}^{\infty} \frac{4 k+3}{4 k+1} \frac{4 k+3}{4 k+5} = \prod_{k=0}^{\infty} \frac{(k+\frac{3}{4})^2}{(k+\frac{1}{4})(k+\frac{5}{4})} = \frac{\Gamma{\left ( \frac{1}{4} \right )} \Gamma{\left ( \frac{5}{4} \right )}}{\Gamma{\left ( \frac{3}{4} \right )^2}} = \frac{2^2 \Gamma{\left ( \frac{5}{4} \right )^2}}{\Gamma{\left ( \frac{3}{4} \right )^2}} $$

The result immediately follows.

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+1. Also, the formula $$\prod_{k=0}^{\infty} \frac{(k+a_1) \cdots (k+a_j)}{(k + b_1) \cdots (k+b_j)} = \frac{\Gamma(b_1) \cdots \Gamma(b_k)}{\Gamma(a_1) \cdots \Gamma(a_k)},$$ where $a_1 + a_2 + \cdots + a_j = b_1 + b_2 + \cdots + b_j$ and no $b_j$ is zero or a negative integer, plus a little manipulation, will finish off your answer. Sondow and Yi give Whittaker and Watson (Section 12.13) as a reference for the formula and state that it is a corollary of the Weierstrass infinite product for the gamma function. –  Mike Spivey Jan 23 '13 at 18:38
    
@MikeSpivey: thanks. Should have dusted off that Whittaker & Watson reference before I typed! –  Ron Gordon Jan 23 '13 at 18:41

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