\(\int_{0}^{\infty}{e^{-x}\ln\bigl({e^x+\sqrt{e^{2x}-1}\,}\bigr) dx} \)
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\(\int_{0}^{\infty}{e^{-x}\ln\bigl({e^x+\sqrt{e^{2x}-1}\,}\bigr) dx} \)
Prove that: \( \displaystyle \int_{0}^{\infty }e^{-x} \ln \left ( e^x+\sqrt{e^{2x}-1} \right )\, dx =\frac{\pi}{2} \).
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Re: \(\int_{0}^{\infty}{e^{-x}\ln\bigl({e^x+\sqrt{e^{2x}-1}\,}\bigr) dx} \)
Hello Apostolos.
Let \(\displaystyle{x\in\left(0,+\infty\right)}\). Then, \(\displaystyle{e^{x}>1}\) and
\(\displaystyle{I(x)=\int_{0}^{x}e^{-t}\,\ln\,\left(e^{t}+\sqrt{e^{2\,t}-1}\right)\,\rm{dt}=\int_{0}^{x}e^{-t}\,\rm{arccosh}\,e^{t}\,\rm{dt}}\) .
We integrate by parts and we get :
\(\displaystyle{\begin{aligned}I(x)&=-\left[\dfrac{\rm{arccosh}\,e^{t}}{e^{t}}\right]_{0}^{x}+\int_{0}^{x}e^{-t}\,\frac{e^{t}}{\sqrt{e^{2\,t}-1}}\,\rm{dt}\\&=-\left[\dfrac{\rm{arccosh}\,e^{t}}{e^{t}}\right]_{0}^{x}+\int_{0}^{x}\dfrac{1}{\sqrt{e^{2\,t}-1}}\,\rm{dt}\end{aligned}}\)
\(\displaystyle{\int_{0}^{x}\dfrac{1}{\sqrt{e^{2\,t}-1}}\,\rm{dt}=\int_{0}^{x}\dfrac{e^{t}}{e^{t}\,\sqrt{e^{2\,t}-1}}\,\rm{dt}}\)
We apply the substitution \(\displaystyle{u=e^{t}\,,u\in\left[1,e^{x}\right]}\) and we have \(\displaystyle{\rm{du}=e^{t}\,\rm{dt}}\) ,so
\(\displaystyle{\int_{0}^{x}\dfrac{1}{\sqrt{e^{2\,t}-1}}\,\rm{dt}=\int_{1}^{e^{x}}\,\dfrac{1}{u\,\sqrt{u^2-1}}\,\rm{du}=\int_{1}^{e^{x}}\,\frac{u}{u^2\,\sqrt{u^2-1}}\,\rm{du}}\)
Now, by substituting \(\displaystyle{s=\sqrt{u^2-1}\,,s\in\left[0,\sqrt{e^{2\,x}-1}\right]}\), we get :
\(\displaystyle{\int_{0}^{x}\dfrac{1}{\sqrt{e^{2\,t}-1}}\,\rm{dt}=\int_{0}^{\sqrt{e^{2\,x}-1}}\dfrac{1}{s^2+1}\,\rm{ds}=\arctan\,\left(\sqrt{e^{2\,x}-1}\right)}\)
Therefore,
\(\displaystyle{\begin{aligned}I(x)&=-\left[\dfrac{\rm{arccosh}\,e^{t}}{e^{t}}\right]_{0}^{x}+\arctan\,\left(\sqrt{e^{2\,x}-1}\right)\\&=-\dfrac{\rm{arccosh}\,e^{x}}{e^{x}}+\arctan\,\left(\sqrt{e^{2\,x}-1}\right)\end{aligned}}\)
where,
\(\displaystyle{\begin{aligned}\lim_{x\to +\infty}\dfrac{\rm{arccosh\,e^{x}}}{e^{x}}&=\lim_{x\to +\infty}\dfrac{\left(\rm{arccosh\,e^{x}}\right)'}{(e^{x})'}\\&=\lim_{x\to +\infty}\dfrac{1}{\sqrt{e^{2\,x}-1}}\\&=0\end{aligned}}\)
and
\(\displaystyle{\lim_{x\to +\infty}\arctan\,\left(\sqrt{e^{2\,x}-1}\right)=\frac{\pi}{2}}\) because
\(\displaystyle{\lim_{x\to +\infty}\sqrt{e^{2\,x}-1}=+\infty}\) .
So,
\(\displaystyle{\lim_{x\to +\infty}I(x)=\frac{\pi}{2}\implies \int_{0}^{\infty}e^{-x}\,\ln\,\left(e^{x}+\sqrt{e^{2\,x}-1}\right)\,\rm{dx}=\frac{\pi}{2}}\) .
Let \(\displaystyle{x\in\left(0,+\infty\right)}\). Then, \(\displaystyle{e^{x}>1}\) and
\(\displaystyle{I(x)=\int_{0}^{x}e^{-t}\,\ln\,\left(e^{t}+\sqrt{e^{2\,t}-1}\right)\,\rm{dt}=\int_{0}^{x}e^{-t}\,\rm{arccosh}\,e^{t}\,\rm{dt}}\) .
We integrate by parts and we get :
\(\displaystyle{\begin{aligned}I(x)&=-\left[\dfrac{\rm{arccosh}\,e^{t}}{e^{t}}\right]_{0}^{x}+\int_{0}^{x}e^{-t}\,\frac{e^{t}}{\sqrt{e^{2\,t}-1}}\,\rm{dt}\\&=-\left[\dfrac{\rm{arccosh}\,e^{t}}{e^{t}}\right]_{0}^{x}+\int_{0}^{x}\dfrac{1}{\sqrt{e^{2\,t}-1}}\,\rm{dt}\end{aligned}}\)
\(\displaystyle{\int_{0}^{x}\dfrac{1}{\sqrt{e^{2\,t}-1}}\,\rm{dt}=\int_{0}^{x}\dfrac{e^{t}}{e^{t}\,\sqrt{e^{2\,t}-1}}\,\rm{dt}}\)
We apply the substitution \(\displaystyle{u=e^{t}\,,u\in\left[1,e^{x}\right]}\) and we have \(\displaystyle{\rm{du}=e^{t}\,\rm{dt}}\) ,so
\(\displaystyle{\int_{0}^{x}\dfrac{1}{\sqrt{e^{2\,t}-1}}\,\rm{dt}=\int_{1}^{e^{x}}\,\dfrac{1}{u\,\sqrt{u^2-1}}\,\rm{du}=\int_{1}^{e^{x}}\,\frac{u}{u^2\,\sqrt{u^2-1}}\,\rm{du}}\)
Now, by substituting \(\displaystyle{s=\sqrt{u^2-1}\,,s\in\left[0,\sqrt{e^{2\,x}-1}\right]}\), we get :
\(\displaystyle{\int_{0}^{x}\dfrac{1}{\sqrt{e^{2\,t}-1}}\,\rm{dt}=\int_{0}^{\sqrt{e^{2\,x}-1}}\dfrac{1}{s^2+1}\,\rm{ds}=\arctan\,\left(\sqrt{e^{2\,x}-1}\right)}\)
Therefore,
\(\displaystyle{\begin{aligned}I(x)&=-\left[\dfrac{\rm{arccosh}\,e^{t}}{e^{t}}\right]_{0}^{x}+\arctan\,\left(\sqrt{e^{2\,x}-1}\right)\\&=-\dfrac{\rm{arccosh}\,e^{x}}{e^{x}}+\arctan\,\left(\sqrt{e^{2\,x}-1}\right)\end{aligned}}\)
where,
\(\displaystyle{\begin{aligned}\lim_{x\to +\infty}\dfrac{\rm{arccosh\,e^{x}}}{e^{x}}&=\lim_{x\to +\infty}\dfrac{\left(\rm{arccosh\,e^{x}}\right)'}{(e^{x})'}\\&=\lim_{x\to +\infty}\dfrac{1}{\sqrt{e^{2\,x}-1}}\\&=0\end{aligned}}\)
and
\(\displaystyle{\lim_{x\to +\infty}\arctan\,\left(\sqrt{e^{2\,x}-1}\right)=\frac{\pi}{2}}\) because
\(\displaystyle{\lim_{x\to +\infty}\sqrt{e^{2\,x}-1}=+\infty}\) .
So,
\(\displaystyle{\lim_{x\to +\infty}I(x)=\frac{\pi}{2}\implies \int_{0}^{\infty}e^{-x}\,\ln\,\left(e^{x}+\sqrt{e^{2\,x}-1}\right)\,\rm{dx}=\frac{\pi}{2}}\) .
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