$\pi$ and Gundermann (happy $\pi$ day, Gauss bell)

$\boxed{\text{1.-}}$ Gundermannian function . Other version of Sahil's solution. Using: $\displaystyle \dfrac{d}{dx} \tanh (x) = \dfrac{d}{dx} \left(\frac{\sinh (x)}{\cosh (x)}\right) = \frac{\cosh^2 (x) - \sinh^2 (x)}{\cosh^2 (x)} = \frac{1}{\cosh^2 (x)}$ $\displaystyle \dfrac{d}{dx} \arcsin (\tanh (x)) = \frac{\tanh ' (x)}{\sqrt{1 - \tanh^2 (x)}} = \frac{\cosh (x)}{\cosh^2 (x)} = \frac{1}{\cosh (x)}$

$\displaystyle \int_{-\infty}^{\infty} \frac{e^{x/2}}{e^x + 1} \,dx = 2 \int_{0}^{\infty} \frac{e^{x/2}}{e^x + 1} \,dx = 2 \int_{0}^{\infty} \frac{1}{e^{(x/2)} + e^{(-x/2)}} \,dx = \int_{0}^{\infty} \frac{1}{ \cosh (x/2)} \, dx =$ Making U- substitution $u = x/2 \to 2 du = dx$ $= 2 \int_{0}^{\infty} \frac{1}{ \cosh (u)} \, du = 2 \arcsin (\tanh(u)) \Big|_{0}^{\infty} = \pi$

$\boxed{\text{2.-}}$ I'll try to develope Chew-Seong and Mark Hennings solutions "a little bit". You can also see page 3 and 4,here where is given $\Gamma(a) \cdot \Gamma (1 - a) = \frac{\pi}{\sin \pi a}, \text{ if } 0 < a < 1$ using substitution for several variables with Jacobian matrix and its determinant.

Let's see first that the above integral converges: Remeber that $0 < a < 1$ \displaystyle \begin{cases}{ \int_{0}^{\infty} \frac{e^{ax}}{e^x + 1} \,dx = \int_{0}^{\infty} \frac{e^{(a - 1)x}}{e^{-x} + 1} \,dx < \int_{0}^{\infty} e^{(a - 1)x} \,dx < + \infty \\ \displaystyle \int_{- \infty}^{0} \frac{e^{ax}}{e^x + 1} \,dx < \int_{- \infty}^{0} e^{ax} \,dx < + \infty} \end{cases} \Rightarrow \displaystyle I = \int_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} \,dx < + \infty

Let $f(x) = \frac{e^{ax}}{e^x + 1}$ , then $e^x + 1 = 0 \iff x \in \{i( \pi + 2k \pi) \text{ / } k \in \mathbb{Z}\}$ , i.e,the poles of $f$ are in $\{i( \pi + 2k \pi) \text{ / } k \in \mathbb{Z}\})$ , i.e, $f \in \mathcal{H}(\mathbb{C} - \{i( \pi + 2k \pi) \text{ / } k \in \mathbb{Z}\}).$ Now, we are going to use Complex analysis and residue theorem .

Let $\gamma_R = \gamma_1 \bigoplus \gamma_2 \bigoplus \gamma_3 \bigoplus \gamma_4$ be the following cycle (see pic), direct sum of piecewise smooth paths, with $R > 0$ very big...

$\gamma_1 (t) = t, \text{, such that } \quad t \in [-R, R]$ .

$\gamma_2 (t) = R + i t \text{, such that } \quad t \in [0, 2 \pi]$ .

$\gamma_3 (t) = t + 2i \pi\text{, such that } \quad t \in [-R , R ]$ .

$\gamma_4 (t) = -R + i t \text{, such that } \quad t \in [0, 2 \pi]$ .

BTW, sorry for the pic, then $\text{Ind(}\gamma_R, i \pi) = 1$ , and applying Residue theorem we get: $\displaystyle I = \int_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} \,dx = \lim_{R \to \infty} \int_{-R}^{R} \frac{e^{ax}}{e^x + 1} \,dx = 2 \pi i \cdot \text{Res(f}, i \pi) \cdot \text{Ind(}\gamma_R, i \pi) = - 2 \pi i \cdot e^{i \pi a}$ since $\text{Res(f}, i \pi) = - e^{i \pi a}$ because $\displaystyle \text{Res(f}, i \pi) = \lim_{z \to i \pi} \frac{e^{az} \cdot (z - i\pi)}{e^z +1} = \lim_{z \to i \pi} \frac{e^{az} \cdot (z - i\pi)}{-(z - i\pi)} = - e^{i \pi a}$ due to power series of $g(z) = e^z$ around $z = i \pi$ . Now, $\displaystyle \oint_{\gamma_R} f(t) \, dt = \oint_{\gamma_1} f(t) \,dt + \oint_{\gamma_2} f(t) \, dt - \oint_{\gamma_3} f(t) \, dt - \oint_{\gamma_4} f(t) \, dt =$ $= \displaystyle \int_{-R}^{R} f(t) \,dt + \int_{0}^{2\pi} f(R + it) \cdot i \,dt - \int_{-R}^{R} f(t + 2\pi i) \,dt - \int_{0}^{2\pi} f(-R + it) \cdot i \,dt = (A)$ On the other hand, $\displaystyle \Big| \oint_{\gamma_2} f(t) \, dt \Big|= \Big| \int_{0}^{2\pi} f(R + it) \cdot i \,dt \Big| = \Big| \int_{0}^{2\pi} \frac{e^{a(R + it)} \cdot i}{1 + e^{R + it}} \,dt \Big| \leq \int_{0}^{2\pi} \frac{e^{aR}}{e^R - 1} \, dt \leq 2\pi \frac{e^{aR}}{e^R - 1} \to 0$ as $R \to +\infty$ , because $| 1 + e^{R + it}| \ge \big| |e^{R + it}| - |-1| \big| = e^R -1, \space (R > 0)$ . In the same way, $\displaystyle \Big| \oint_{\gamma_4} f(t) \, dt \Big| \to 0, \text{ as } R \to \infty.$ Therefore, $\displaystyle \oint_{\gamma_R} f(t) \, dt = \oint_{\gamma_1} f(t) \,dt - \oint_{\gamma_3} f(t) \, dt = \int_{-R}^{R} \frac{e^{at}}{e^t + 1} \,dt - \int_{-R}^{R} \frac{e^{a(t + 2i \pi)}}{e^{t + 2i \pi} + 1} \,dt$ and $\displaystyle - \int_{-R}^{R} \frac{e^{a(t + 2i \pi)}}{e^{t + 2i \pi} + 1} \,dt = - e^{a2\pi i} \cdot \int_{-R}^{R} \frac{e^{at}}{e^t + 1} \,dt \Rightarrow$ $I(1 - e^{a2\pi i}) = - 2\pi i \cdot e^{\pi i a} \Rightarrow I = \frac{- 2\pi i \cdot e^{\pi i a}}{1 - e^{a2\pi i}} = \frac{\pi}{\sin \pi a}$

$\boxed{\text{q. e. d}}$

$\begin{aligned} I(a) & = \int_{-\infty}^\infty \frac {e^{ax}}{e^x+1} dx & \small \color{#3D99F6} \text{Let }u = e^x, \ du = e^x dx \\ & = \int_0^\infty \frac {u^{a-1}}{u+1} dx & \small \color{#3D99F6} \text{Converting to beta function:} \\ & = B(a, 1-a) & \small \color{#3D99F6} B(m+1, n-m-1) = \int_0^\infty \frac {u^m}{(1+u)^n} du \end{aligned}$

$\begin{aligned} \implies I \left( \frac 12 \right) & = B \left(\frac 12, 1 - \frac 12 \right) \\ & = B \left(\frac 12, \frac 12 \right) \\ & = \frac {\Gamma\left(\frac 12 \right) \Gamma\left(\frac 12 \right)}{\Gamma\left( 1\right)} & \small \color{#3D99F6} \Gamma(\cdot) \text{ denotes the gamma function} \\ & = \frac {\sqrt \pi \cdot \sqrt \pi}{0!} \\ & = \pi \approx \boxed{3.141592} \end{aligned}$

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References:

• Beta function
• Gamma function

$\large \color{#E81990} Happy \ \huge \pi \large \ Day$

We can deal this problem with U-substitution as follows, $\displaystyle \int_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} \,dx$ So, let $u=e^{x/2}$ , $\frac{\text{d}u}{\text{dx}}=\frac{e^{x/2}}{2}$ $\text{d}x=\frac{2\text{d}u}{u}$ By substituting $e^{x/2}=u$ , $\displaystyle \int_{0}^{\infty} \frac{u}{u^2 + 1} \frac{2du}{u}$ $\displaystyle \int_{0}^{\infty} \frac{2}{u^2 + 1} \,du=2\arctan(u)=2\arctan(e^{x/2})=\pi$

Another way:

We have an integral, $\displaystyle \int_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} \,dx$ $\displaystyle \int_{-\infty}^{\infty} \frac{ \text{sech}(\frac{x}{2})}{2} \,dx$ Let's substitute $u=\frac{x}{2}$ , $\displaystyle \frac{du}{dx}=\frac{1}{2}$ $\displaystyle dx=2*du$ $\displaystyle \int_{-\infty}^{\infty} \text{sech(u)} \ du$ By rewriting we get, $\displaystyle \int_{-\infty}^{\infty} \frac{\text{cosh(u)}}{\sinh^2(u)+1} du$ Let's substitute $v=\sinh(u)$ , $\displaystyle \frac{dv}{du}=\cosh(u)$ $\displaystyle dv=du*\cosh(u)$ $\displaystyle \int_{-\infty}^{\infty} \frac{1}{v^2+1} \ dv=\arctan(u)=arctan\left(\sinh\left(\frac{x}{2}\right)\right)=\pi$

Let $\gamma_1$ be the line segment $[-X,X]$ , let $\gamma_2$ be the line segment $[-X+2\pi i,X+2\pi i]$ , let $\gamma_3$ be the line segment $[X,X+2\pi i]$ , and let $\gamma_4$ be the line segment $[-X,-X+2\pi i]$ for any $X > 0$ . Then, for $0 < a < 1$ , $\left(\int_{\gamma_1} + \int_{\gamma_3} - \int_{\gamma_2} - \int_{\gamma_4}\right)\frac{e^{az}}{e^z+1}\,dz \; = \; 2\pi i \mathrm{Res}_{z=\pi i}\frac{e^{az}}{e^z+1} \; = \; 2\pi i \frac{e^{\pi i a}}{-1} \; =\; -2\pi i e^{\pi i a}$ But $\int_{\gamma_1} \frac{e^{az}}{e^z+1}\,dz \; = \; \int_{-X}^X \frac{e^{ax}}{e^x+1}\,dx \hspace{2cm} \int_{\gamma_2} \frac{e^{az}}{e^z+1}\,dz \; = \; e^{2\pi i a}\int_{-X}^X \frac{e^{ax}}{e^x+1}\,dx$ while $\int_{\gamma_3}\frac{e^{az}}{e^z+1}\,dz \; = \; O(e^{-(1-a)X}) \hspace{2cm} \int_{\gamma_4}\frac{e^{az}}{e^z+1}\,dz \; = \; O(e^{-aX})$ as $X \to \infty$ . Letting $X \to \infty$ , we have $(1 - e^{2\pi i a})\int_{-\infty}^\infty \frac{e^{ax}}{e^x+1}\,dx \; =\; -2\pi i e^{\pi i a}$ and so $\int_{-\infty}^\infty \frac{e^{ax}}{e^x+1}\,dx \; =\; \frac{\pi}{\sin \pi a}$

$I=\displaystyle \int_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} \,dx= \int_{-\infty}^{0} \frac{e^{ax}}{e^x + 1} \,dx+ \int_{0}^{\infty} \frac{e^{ax}}{e^x + 1} \,dx=I_1+I_2$

$I_1= \displaystyle\int_{-\infty}^{0} \frac{e^{ax}}{e^x + 1} \,dx= \int_{-\infty}^{0}\sum_{n=0}^{\infty} (-1)^n e^{(a+n)x}\,dx=\sum_{n=0}^{\infty} \frac{(-1)^n}{n+a}=\frac{1}{a}+\sum_{n=1}^{\infty} \frac{(-1)^n}{n+a}$

Here, we have taken $\displaystyle\frac{1}{1+e^x}=\sum_{n=0}^{\infty} (-1)^n e^{nx}$ as $e^x<1$ for $x<0$ .

$I_2= \displaystyle\int_{0}^{\infty} \frac{e^{ax}}{e^x + 1} \,dx= \sum_{n=0}^{\infty} (-1)^n e^{(a-n-1)x}\,dx=\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1-a}=\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m-a}$

In this case, $e^x>1$ and $e^{-x}<1$ as $x>0$ . So, we take $\displaystyle\frac{1}{1+e^x}=\frac{e^{-x}}{1+e^{-x}}=\sum_0^{\infty} (-1)^n e^{-(n+1)x}$ .

Again, $\displaystyle\lim_{x\rightarrow\infty}e^{(a-n-1)x}=0$ for $0<a<1$ and $n\geq0$ . $\implies I=\frac{1}{a}+\sum_{n=1}^{\infty} \frac{(-1)^n}{n+a}+\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m-a} \\ =\frac{1}{a}+2a\sum_{n=1}^{\infty} \frac{(-1)^n}{a^2-n^2}\\ =\pi\bigg[\frac{1}{a\pi}+2a\pi\sum_{n=1}^{\infty} \frac{(-1)^n}{a^2\pi^2-n^2\pi^2}\bigg]$

From the pole expansion of meromorphic function, we know that $\csc z=\frac{1}{\sin z}=\frac{1}{z}+2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2 \pi^2}$ $\implies I=\frac{\pi}{\sin (\pi a)}$

For a=1/2 , let (e^x +1)=t and solve