# How to evaluate $\int_{0}^{\infty}e^{-x}f\left(x\right)dx$ using $f\left(0\right)$ and $f\left(1\right)$?

Let $$f\in{\cal C}^{\infty}\left(\mathbb{R}\right)$$ be such that $$\left|f^{\left(k\right)}\left(x\right)\right|\leq M_{k}$$ for every $$k$$. Find a way to evaluate $$\int_{0}^{\infty}e^{-x}f\left(x\right)dx$$ using $$f\left(0\right)$$ and $$f\left(1\right)$$, and find an expression for the error.

So I used the fact that $$\int_{0}^{\infty}e^{-x}f\left(x\right)dx=\int_{0}^{1}e^{-x}f\left(x\right)dx+\int_{1}^{\infty}e^{-x}f\left(x\right)dx$$ And by the trapezoid rule \begin{align*} \int_{0}^{1}e^{-x}f\left(x\right)dx & =\frac{\left(e^{-0}f\left(0\right)+e^{-1}f\left(1\right)\right)}{2}-\frac{1}{12}g''\left(\xi\right)=\\ & =\frac{1}{2}\left(f\left(0\right)+\frac{f\left(1\right)}{e}\right)-\frac{1}{12}g''\left(\xi\right) \end{align*} where $$g\left(x\right)=e^{-x}f\left(x\right)$$. Now as for the second integral I tried to use the substitution $$x=\frac{1}{t},dx=-\frac{dt}{t^{2}}$$ so $$\int_{1}^{\infty}e^{-x}f\left(x\right)dx=-\int_{1}^{0}\frac{e^{-\frac{1}{t}}f\left(\frac{1}{t}\right)}{t^{2}}dt=\int_{0}^{1}\frac{e^{-\frac{1}{t}}f\left(\frac{1}{t}\right)}{t^{2}}dt$$ But how how to continue from here? Any help?

• Can't you do the same? How accurate is the error meant to be? – Fimpellizieri Jan 17 at 18:17
• @Jon If you try quadratures of the form $Q(f) = A f(0) + B f(1)$, you will see that they can only have degree 1 if $A=0$ and $B=1$. So, in terms of degree, the best you can do is to set $Q(f)=f(1)$. – PierreCarre Jan 20 at 11:25
• @PierreCarre Can you please elaborate how to get this? Maybe as an answer? – Jon Jan 21 at 7:40

This is a partial answer. If you assume that your quadrature rule is of the form $$Q(f)= A f(0) + B f(1)$$ and force it to be exact for polynomials of degree $$\leq 1$$, you get $$Q(1) = \int_0^{+\infty} e^{-x} \cdot 1 dx, \quad Q(x)=\int_0^{+\infty} e^{-x} \cdot x dx,$$
where $$Q(1)= A + B, \quad Q(x) = A \cdot 0 + B \cdot 1 = B.$$
This leads you to the conclusion that the quadrature can only have degree one if $$A+B=1$$ and $$B=1$$, i.e. $$A=0$$ and $$B=1$$. So, the only quadrature rule of the proposed type that has degree one in simply $$Q(f) = f(1)$$. Regarding the error, we can for instance see that \begin{align*} \int_0^{+\infty} & f(x) e^{-x} dx - Q(f) = \int_0^{+\infty}(f(x)-f(1))e^{-x} dx \\ & = \int_0^{+\infty}(f(1) + f'(1)(x-1) + \frac 12 f''(\xi_x)(x-1)^2-f(1))e^{-x} dx\\ & =\frac 12 \int_0^{+\infty}f''(\xi_x) (x-1)^2 e^{-x} dx = \frac 12 f''(\xi)\\ & \end{align*}
• Why can we assume $Q(f)= A f(0) + B f(1)$ is the form of the rule? – Jon Jan 21 at 11:07
• @Jon Quadrature rules are (almost) always of the form $Q(f)=\sum_{i=0}^n \omega_i f(x_i)$, where $x_i$ are the quadrature nodes and $\omega_i$ are the quadrature weights. Since we are supposed to use $f(0)$ and $f(1)$... This particular form of the quadrature rules yields a linear application, just like the integral. – PierreCarre Jan 21 at 13:25