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Look-alike integrals

$$I_1+I_2=\int_{-\infty}^{\infty}{e^x+1\over (e^x-x+1)^2+\pi^2}\mathrm dx=\int_{-\infty}^{\infty}{e^x+1\over (e^x+x+1)^2+\pi^2}\mathrm dx=1\tag1$$

I just wonder if these integrals $I_1$ and $I_2$ are the same in term of transforming is concern? Or they just only happened to give the same closed form?

If I make a substitution of $u=e^x+1$, nothing much happened

$$\int_{1}^{\infty}{u\over u-1}\cdot{\mathrm du\over (u-\ln{(u-1)})^2+\pi^2}=\int_{1}^{\infty}{u\over u-1}\cdot{\mathrm du\over (u+\ln{(u-1)})^2+\pi^2}\tag2$$

Note: I can't show it but I think $I_1$ and $I_2$ are not related in term of transforming into each other.

Else

How can we show that $I_1$ or $I_2$ has a value of one?

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    $\begingroup$ we had this kind of integrals before, i remember that @robjohn once wrote an excellent post about them $\endgroup$ – tired Jan 25 '17 at 11:46
  • $\begingroup$ Guess we must use the fact that both functions have nice poles at $x=\pm\pi i$. $\endgroup$ – Ivan Neretin Jan 25 '17 at 11:55
  • $\begingroup$ Thank you @tired I check it now $\endgroup$ – gymbvghjkgkjkhgfkl Jan 25 '17 at 11:57
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    $\begingroup$ Such integrals appear to be related with Gregory coefficients: en.wikipedia.org/wiki/Gregory_coefficients $\endgroup$ – Jack D'Aurizio Jan 25 '17 at 12:11
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    $\begingroup$ And that relation can be hinted from the complex-analytic proof. For instance, here is (German-version of) a proof of related integrals. (See the item 2.5 and then click 'Beweis' to expand the proof.) Certainly a small tweak will give you a proof for that identity. $\endgroup$ – Sangchul Lee Jan 25 '17 at 12:15
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We have that $\frac{e^x-1}{(e^x-x+1)^2+\pi^2}$ has a simple primitive, given by $\frac{1}{\pi}\arctan\frac{1+e^x-x}{\pi}$. It follows that:

$$ I_1 = \int_{-\infty}^{+\infty}\frac{e^x+1}{(e^x-x+1)^2+\pi^2}\,dx= 2\int_{-\infty}^{+\infty}\frac{dx}{(e^x-x+1)^2+\pi^2}$$ and the residue theorem gives the following

Lemma. If $a>0$ and $b\in\mathbb{R}$, $$ \int_{-\infty}^{+\infty}\frac{a^2\,dx}{(e^x-ax-b)^2+(a\pi)^2}=\frac{1}{1+W\left(\frac{1}{a}e^{-b/a}\right)} $$ where $W$ is Lambert's function.

In our case, by choosing $a=1$ and $b=-1$ we get that $I_1$ depends on $W(e)=1$ and equals $\color{red}{\large 1}$.

$I_2$ is easier to compute: $$ I_2 = \int_{-\infty}^{+\infty}\frac{e^x+1}{(e^x+x+1)^2+\pi^2}\,dx = \frac{1}{\pi}\,\left.\arctan\left(\frac{1+x+e^x}{\pi}\right)\right|_{-\infty}^{+\infty} = \color{red}{1}.$$


Addendum. Due to the identity $$\begin{eqnarray*} I_1 &=& \int_{0}^{+\infty}\frac{u+1}{u}\cdot\frac{du}{(u+1-\log u)^2+\pi^2}\\&=&\frac{1}{\pi}\int_{0}^{+\infty}\int_{0}^{+\infty}(u+1)u^{x-1}e^{-(u+1)x}\sin(\pi x)\,dx\,du\\&=&\frac{2}{\pi}\int_{0}^{+\infty} e^{-x} x^{-x}\Gamma(x)\sin(\pi x)\,dx\\&=&2\int_{0}^{+\infty}\frac{e^{-x} x^{-x}}{\Gamma(1-x)}\,dx\end{eqnarray*}$$ the previous Lemma also proves the highly non-trivial identity $$ \int_{0}^{+\infty}\frac{(ex)^{-x}}{\Gamma(1-x)}\,dx = \frac{1}{2} \tag{HNT}$$ equivalent to the claim $I_1=1$. It would be interesting to find an independent proof of $\text{(HNT)}$, maybe based on Glasser's master theorem, Ramanujan's master theorem or Lagrange inversion. There also is an interesting discrete analogue of $\text{(HNT)}$, $$ \sum_{n\geq 1}\frac{n^n}{n!(4e)^{n/2}}=1$$ that comes from the Lagrange inversion formula.

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