Find \(\int_{0}^{\infty} \mathrm{e}^{-x} \cos x \mathrm{~d} x\).

Short Answer

Expert verified
Based on the given solution, the improper integral \(\int_{0}^{\infty} \mathrm{e}^{-x} \cos x \mathrm{~d} x\) equals to \(\frac{1}{2}\).

Step by step solution

01

Identify u and dv

We need to choose the differential d(u) and the function v in the integration by parts. In this case, we will set: u = cos(x), d(u) = -sin(x) dx dv = e^(-x) dx, v = -e^(-x)
02

Apply integration by parts

We apply the formula \(\int u \mathrm{d}v = uv - \int v \mathrm{d}u\). Our integral becomes: \(-\int_{0}^{\infty} \cos x \cdot e^{-x} dx = -\cos(x)e^{-x} \Big|_{0}^{\infty} + \int_{0}^{\infty} e^{-x} (\sin x) dx\)
03

Evaluate the term at the limits

We need to evaluate \(-\cos(x)e^{-x} \Big|_{0}^{\infty}\): \(- \lim_{x \to \infty} (\cos(x) e^{-x}) + (\cos(0) e^{0}) = 0 + 1 = 1\)
04

Apply integration by parts again for the remaining integral

We now need to integrate \(\int_{0}^{\infty} e^{-x} (\sin x) dx\) using integration by parts. Choosing: u = sin(x), d(u) = cos(x) dx dv = e^(-x) dx, v = -e^(-x) Now apply the integration by parts formula: \(\int_{0}^{\infty} e^{-x} (\sin x) dx = - e^{-x}\sin(x) \Big|_{0}^{\infty} - \int_{0}^{\infty} - e^{-x} (\cos x) dx\)
05

Evaluate the term at the limits and solve for remaining integral

We need to evaluate \(- e^{-x}\sin(x) \Big|_{0}^{\infty}\): \(- \lim_{x \to \infty} (e^{-x} \sin(x)) - (e^{0} \sin(0)) = 0\) Thus, we have: \(\int_{0}^{\infty} e^{-x} (\sin x) dx = \int_{0}^{\infty} - e^{-x} (\cos x) dx\)
06

Combine terms and solve for the original integral

Now we can express the original integral as: \(\int_{0}^{\infty} e^{-x} \cos x dx = 1 + \int_{0}^{\infty} e^{-x} (\sin x) dx = 1 - \int_{0}^{\infty} e^{-x} (\cos x) dx\) Rearrange the equation to isolate the integral: \(2\int_{0}^{\infty} e^{-x} \cos x dx = 1\) \(\int_{0}^{\infty} e^{-x} \cos x dx = \frac{1}{2}\)

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