\( \int_{0}^{1}{\rm erf^2}(x)\,dx \)
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
\( \int_{0}^{1}{\rm erf^2}(x)\,dx \)
Evaluate : \( \displaystyle \int_{0}^{1}{\rm erf^2}(x)\,dx \).
Imagination is much more important than knowledge.
-
- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Re: \( \int_{0}^{1}{\rm erf^2}(x)\,dx \)
Hello Tolis.
The function \(\displaystyle{\rm{erf}}\) is well defined, continuous and analytic on \(\displaystyle{\mathbb{C}}\) and given
by \(\displaystyle{\rm{erf}(z)=\dfrac{2}{\sqrt{\pi}}\,\int_{0}^{z}e^{-t^2}\,\mathrm{d}t}\) , cause the function \(\displaystyle{t
\mapsto e^{-t^2}\,,t\in\mathbb{C}}\)
is continuous and analytic on \(\displaystyle{\mathbb{C}}\) , with \(\displaystyle{\left(\rm{erf}(z)\right)'=\dfrac{2}{\sqrt{\pi}}\,e^{-z^2}\,,z\in\mathbb{C}}\) .
By applying integration by parts, we get :
\(\displaystyle{\begin{aligned} \int_{0}^{1}\rm{erf}^2\,(x)\,\mathrm{d}x&=\left[x\,\rm{erf}^2\,(x)\right]_{0}^{1}-\frac{4}{\pi}\int_{0}^{1}2\,x\,\rm{erf}\,(x)\,\left(\rm{erf}\,(x)\right)'\,\mathrm{d}x\\&=\rm{erf}^2\,(1)-\frac{4}{\pi}\int_{0}^{1}\rm{erf}\,(x)\,\mathrm{d}\,(e^{-x^2})\\&=\rm{erf}^2\,(1)-\frac{4}{\pi}\left[\rm{erf}\,(x)\,e^{-x^2}\right]_{0}^{1}+\frac{4}{\pi}\int_{0}^{1}e^{-2\,x^2}\,\mathrm{d}x\\&=\rm{erf}^2\,(1)-\frac{4}{\pi}e^{-1}\,\rm{erf}\,(1)+\sqrt{\frac{2}{\pi}}\int_{0}^{1}e^{-(\sqrt{2}\,x)}\,\mathrm{d}x\\&=\rm{erf}^2\,(1)-\frac{4}{\pi}e^{-1}\,\rm{erf}\,(1)+\sqrt{\frac{2}{\pi}}\rm{erf}\,(\sqrt{2})\end{aligned}}\)
The function \(\displaystyle{\rm{erf}}\) is well defined, continuous and analytic on \(\displaystyle{\mathbb{C}}\) and given
by \(\displaystyle{\rm{erf}(z)=\dfrac{2}{\sqrt{\pi}}\,\int_{0}^{z}e^{-t^2}\,\mathrm{d}t}\) , cause the function \(\displaystyle{t
\mapsto e^{-t^2}\,,t\in\mathbb{C}}\)
is continuous and analytic on \(\displaystyle{\mathbb{C}}\) , with \(\displaystyle{\left(\rm{erf}(z)\right)'=\dfrac{2}{\sqrt{\pi}}\,e^{-z^2}\,,z\in\mathbb{C}}\) .
By applying integration by parts, we get :
\(\displaystyle{\begin{aligned} \int_{0}^{1}\rm{erf}^2\,(x)\,\mathrm{d}x&=\left[x\,\rm{erf}^2\,(x)\right]_{0}^{1}-\frac{4}{\pi}\int_{0}^{1}2\,x\,\rm{erf}\,(x)\,\left(\rm{erf}\,(x)\right)'\,\mathrm{d}x\\&=\rm{erf}^2\,(1)-\frac{4}{\pi}\int_{0}^{1}\rm{erf}\,(x)\,\mathrm{d}\,(e^{-x^2})\\&=\rm{erf}^2\,(1)-\frac{4}{\pi}\left[\rm{erf}\,(x)\,e^{-x^2}\right]_{0}^{1}+\frac{4}{\pi}\int_{0}^{1}e^{-2\,x^2}\,\mathrm{d}x\\&=\rm{erf}^2\,(1)-\frac{4}{\pi}e^{-1}\,\rm{erf}\,(1)+\sqrt{\frac{2}{\pi}}\int_{0}^{1}e^{-(\sqrt{2}\,x)}\,\mathrm{d}x\\&=\rm{erf}^2\,(1)-\frac{4}{\pi}e^{-1}\,\rm{erf}\,(1)+\sqrt{\frac{2}{\pi}}\rm{erf}\,(\sqrt{2})\end{aligned}}\)
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 3 guests