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I've been trying to solve this integral for a few days.

$$\int_0^{\infty}\left(\frac{1}{n}\left(t+n\right)\ln\left(\frac{t+n}{t}\right)-\ln\left(\frac{1}{n}\left(t+n\right)\ln\left(\frac{t+n}{t}\right)\right)-1\right)dt$$

For $n\gt0$.

I'm able to solve most of the integral until I got stuck trying to solve

$$\int\log\left(\log\left(\frac{t+n}{t}\right)\right)dt$$

Edit: We first see that with a substitution we take $n$ out of the problem. Thus the integral we want to solve has a value of $0.38033\dots$ @Yuriy S has helped find an alternate form for integral. I want to contribute another alternate form that can be derived from Yuriy's form. $$I_1=-\frac{1}{4}+\int_{0}^{\infty}\left(-\frac{e^s-1}{2e^s}+\ln\left(e^s-1\right)-\ln\left(s\right)\right)\frac{e^s}{\left(e^s-1\right)^2}ds$$

Another Update: I discovered that

$$\begin{align} I_1+\frac14&=-\int_x^\infty\frac{1}{t(e^t-1)}dt-\left(-\frac1x-\frac{\ln{x}}{2}+\sum_{n=2}^\infty\frac{B_n}{n!(n-1)}x^{n-1}\right) \\&=\sum_{n=1}^\infty \text{Ei}(-xn)-\left(-\frac1x-\frac{\ln{x}}{2}+\sum_{n=2}^\infty\frac{B_n}{n!(n-1)}x^{n-1}\right) \end{align}$$ for $0\lt x\lt 2\pi$. Here the non-integral part on the rhs is the series expansion of the integral part at $x=0$.

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  • $\begingroup$ Numerically, the value of the integral seems to be about $0.38033n$. $\endgroup$ – MathIsFun7225 Aug 12 '19 at 14:23
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Simplify first by substitution:

$$t=nu$$

$$I(n)=n\int_0^{\infty}\left((u+1)\ln\left(1+\frac{1}{u}\right)-\ln\left((u+1)\ln\left(1+\frac{1}{u}\right)\right)-1\right)du$$

So we need to find the following constant:

$$I_1=\int_0^{\infty}\left((u+1)\ln\left(1+\frac{1}{u}\right)-\ln\left((u+1)\ln\left(1+\frac{1}{u}\right)\right)-1\right)du$$

Frankly, this looks bad, I doubt there's an exact solution. The numerical value, as stated by MathIsFun7225 is about $0.3803301$.

Using some substitutions, we can transform the integral to:

$$I_1=\int_0^{\infty}\left(\frac{s}{e^s-1}+\ln(e^s-1)-\ln s-1\right)\frac{e^s ds}{(e^s-1)^2}$$

The function:

$$f(s)=s+(e^s-1)\left(\ln(e^s-1)-\ln s-1\right)$$

Has a nice Taylor expansion around zero:

$$f(s)= \frac{s^3}{8}+\frac{s^4}{16}+\frac{11s^5}{576}+\frac{5s^6}{1152}+\frac{41s^7}{51840}+\frac{5s^8}{41472}+\dots \tag{1}$$

$$I_1= \int_0^{\infty}f(s)\frac{e^s ds}{(e^s-1)^3} \tag{2}$$

Consider:

$$J_k=\int_0^{\infty}\frac{s^k e^s ds}{(e^s-1)^3}$$

In the answer(s) to this question: An integral for the difference of zeta functions $\zeta (s-1)-\zeta(s)$ it is shown that:

$$J_k=\frac{k!}{2} (\zeta(k-1)-\zeta(k)) \tag{3}$$


Finally, summing a few first terms of the series (1), we obtain a number close to numerical value of the integral.

For example three first terms give us the value $0.3079 \ldots$.

The first six terms give $0.3668 \dots$.

However, since the Taylor series (1) has a finite radius of convergence, the series obtained for the integral is asymptotic in nature and most likely diverges. But as usual with asymptotic series, finitely many terms should give a good approximation for the integral.

The series terms all have the same sign up to $s^{20}$, then we encounter the first sign change. Summing all the terms for $k=3, \dots, 20$, we obtain:

$$I_1 \approx 0.3803246 \dots$$

Which is a good approximation. I'm not sure what number of terms will give the best agreement with the exact value.


Update:

Using @automaticallyGenerated's answer, I have checked the asymptotic series numerically, and here's the result for different number of terms (starting with $k=3$:

$$\left( \begin{array}{cc} 15 & 0.380130074058105238689754781268 \\ 16 & 0.380223929458113985169381973291 \\ 17 & 0.380272711854003260001162359969 \\ 18 & 0.380298890136158789781977313273 \\ 19 & 0.380315792727508660246473419445 \\ 20 & 0.380324694728276221658188520931 \\ 21 & 0.380323061796211720843375783973 \\ 22 & 0.380322359261040332671841945024 \\ 23 & 0.380338844375899979977446596772 \\ 24 & 0.380347115006252429034820349587 \\ 25 & 0.380297504104854694212803005034 \\ 26 & 0.380272706474517002755450053827 \\ 27 & 0.380464567324088204449788570410 \\ 28 & 0.380560498418357337146304250784 \\ 29 & 0.379700469605659832230621914362 \\ 30 & 0.379270457893190348577342716132 \\ 31 & 0.383724367550841791140185335387 \\ 32 & 0.385951317329242809053234753014 \\ 33 & 0.359568287367679813690910184314 \\ 34 & 0.346376786006546776686420830652 \\ 35 & 0.523655288520764751498176067439 \\ 36 & 0.612294499198448476853004251809 \\ 37 & -0.72901761727886247456922591902 \\ 38 & -1.39967353904013645644838768421 \\ 39 & 9.9527821590780141647092153950 \\ 40 & 15.6290094947770368344784462912 \\ 41 & -91.221244446733759055679027656 \\ 42 & -144.646369270054172239030454206 \\ 43 & 967.80090283422936072262158677 \\ 44 & 1524.02452894976299543567171838 \\ 45 & -11226.0112127462193171191333164 \\ \end{array} \right)$$

Clearly, the series doesn't converge and the best approximation is achieved around $k_m=20$.

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  • $\begingroup$ Why don't you use $$f(x)=\frac{s}{e^s-1}+\ln(e^s-1)-\ln s-1$$ and $$J_k = \int_0^\infty \frac{s^k e^s ds}{(e^s-1)^2} = k! \cdot \zeta(k)$$? It seems like it would be easier to approximate that way. $\endgroup$ – VVejalla Aug 22 '19 at 16:58
  • $\begingroup$ @automaticallyGenerated, I get rid of the logarithm that way, after series expansion. But your way is good as well $\endgroup$ – Yuriy S Aug 22 '19 at 17:04
  • $\begingroup$ I believe we have $\frac{s}{e^s-1}+\ln(e^s-1)-\ln(s)-1=\sum_{n=1}^\infty{\frac{(2n+1)B_{2n}}{(2n)!(2n)}s^{2n}}$ for $|s|\lt 2\pi$ $\endgroup$ – tyobrien Aug 23 '19 at 5:13
  • $\begingroup$ @automaticallyGenerated, regarding your answer, the obvious problem is that the series for $f(s)$, either mine or yours has a finite radius of convergence, which means that termwise integration can only give us an asymptotic series. Good for approximation, but diverges $\endgroup$ – Yuriy S Aug 23 '19 at 8:34
  • $\begingroup$ Right, do you know of any techniques to analytically continue the domain of the series? $\endgroup$ – tyobrien Aug 23 '19 at 11:34
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We can actually find a term by term expansion of the $f(s)=s+(e^s-1)\left(\ln(e^s-1)-\ln s-1\right)$ used in YuriyS's answer. If we rearrange $f(s)$, we get $$f(x) = (x+1-e^x) + (e^x-1)(\ln(e^x-1)-\ln(x))$$

If we look at $x+1-e^x$, this has a known Taylor series (which converges for all real) of $$-\sum_{n=2}^\infty \frac{x^n}{n!}$$

We also know that $$e^x-1 = \sum_{n=1}^\infty \frac{x^n}{n!}$$ which again converges for all real.

$\ln(e^x-1)-\ln(x)$ is a bit more tricky. If we differentiate it, we get $$\frac{e^x}{e^x-1}-\frac{1}{x} = 1+\frac{1}{e^x-1}-\frac{1}{x}$$ Here we can use the fact that $\frac{x}{e^x-1} = \sum_{n=0}^\infty \frac{B_n}{n!} x^n$ where $B_n$ are the Bernoulli numbers. If we divide by $x$ and add $1-\frac{1}{x}$, we get $$1+\frac{1}{e^x-1}-\frac{1}{x} = \frac{1}{2}+\sum_{n=2}^{\infty}\frac{B_n}{n!} x^{n-1}$$ Integrating, we then have that $$\ln(e^x-1)-\ln(x) = \frac{x}{2} + \sum_{n=2}^\infty\frac{B_n}{n! \cdot n}x^n$$

We now have that $$f(x) = -\sum_{n=2}^\infty \frac{x^n}{n!} + \sum_{n=1}^\infty \frac{x^n}{n!} \cdot \left(\frac{x}{2} + \sum_{n=2}^\infty\frac{B_n}{n! \cdot n}x^n\right) = -\sum_{n=2}^\infty \frac{x^n}{n!} + \frac{x}{2} \sum_{n=1}^\infty \frac{x^n}{n!} + \sum_{n=1}^\infty \frac{x^n}{n!} \cdot \sum_{m=2}^\infty\frac{B_m}{m! \cdot m}x^m$$

$$f(x) = -\sum_{n=2}^\infty \frac{x^n}{n!} + \sum_{n=2}^\infty \frac{x^{n}}{2(n-1)!} + \sum_{n=3}^\infty \cdot \sum_{m=2}^{n-1}\frac{B_m}{(n-m)!m! \cdot m}x^n$$

Finally, we get the closed form for $a_n$ in $f(x) = \sum_{n=3}^\infty a_n x^n$ as $$a_n = \frac{n-2}{2(n!)}+\sum_{m=2}^{n-1}\frac{B_m}{(n-m)!m! \cdot m} = \sum_{m=2}^{n-1}\left(\frac{B_m}{(n-m)!m! \cdot m}+\frac{1}{2(n!)}\right)$$

Using the same $I_1 = \int_0^\infty f(s) \frac{e^s ds}{(e^s-1)^3}$ as YuriyS, we now want to find $$I_1 = \sum_{n=3}^\infty a_n \frac{n!}{2} (\zeta(n-1)-\zeta(n)) = \sum_{n=2}^\infty \left(a_{n+1}\frac{(n+1)!}{2}-a_{n}\frac{n!}{2}\right)\zeta(n)$$ I am not exactly sure what to do from here, but at least it is in the form of an infinite series instead of an integral.

Edit: As YuriyS mentioned in the comments, $a_n n!$ can be stated neatly as $b_n=\frac{n-2}{2}+\sum_{m=2}^{n-1}\frac{B_m \binom{n}{m}}{m}$. This means that $$I_1 = \sum_{n=3}^\infty \frac{b_n}{2} (\zeta(n-1)-\zeta(n)) = \sum_{n=2}^\infty \left(\frac{b_{n+1}}{2}-\frac{b_n}{2}\right)\zeta(n)$$

Edit 2: Unfortunately, these series will diverge, as mentioned in the comments, making it impossible for use in calculating $I_1$.

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  • $\begingroup$ You can rewrite $a_n n!$ more neatly using $\binom{n}{m}$ $\endgroup$ – Yuriy S Aug 23 '19 at 17:41
  • $\begingroup$ The final series doesn't converge. See my edit. It's still an asymptotic series. Sometimes even when the series inside the integral converges, the termwise integration still leads to divergent series $\endgroup$ – Yuriy S Aug 23 '19 at 18:00
  • $\begingroup$ Though in this case, why are you so sure that $\frac{x}{e^x-1} = \sum_{n=0}^\infty \frac{B_n}{n!} x^n$ converges for all real $x$? Numerical experiments suggest otherwise, as for large $x$ the terms start to grow in absolute value $\endgroup$ – Yuriy S Aug 23 '19 at 18:04

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