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Evaluate  \(\int_{-\infty}^{0} x e^{x} d x\)

\(\begin{array}{c}{\int_{-\infty}^{0} x e^{x} d x=\lim _{t \rightarrow-\infty} \int_{t}^{0} x e^{x} d x}\end{array}\)

Using integration by parts 

let \(u=x \quad \Rightarrow d u=1 d x\)

\(v=e^{x} \quad \Leftarrow d v=e^{x} d x\)

\(\int_{t}^{0} x e^{x} d x=u \cdot\left.v\right|_{t} ^{0}-\int_{t}^{0} v d u\)

\(=x\left.e^{x}\right|_{t} ^{0}-\int_{t}^{0} e^{x} \cdot d x\)

\(=\left(0-t e^{t}\right)-\left.e^{x}\right|_{t}^{0}\)

\(=-te^{t}-\left(e^{0}-e^{t}\right)=-t e^{t}-1+e^{t}\)

\(\int_{t}^{0} x e^{x} d x=\lim _{t \rightarrow-\infty}\left[-t e^{t}-1+e^{t}\right]\)

\(\begin{array}{l}{=-\lim _{t \rightarrow-\infty} te^{t}-\lim_{t \rightarrow-\infty} (1)+\lim _{t \rightarrow-\infty} e^{t}=-\lim _{t \rightarrow-\infty} te^{t}-1+0}\end{array}\)

\(-\lim _{t \rightarrow-\infty} t e^{t}=-\left(-\infty e^{-\infty}\right)=-(-\infty \cdot(0))=\infty(0)\)

use l'hoptal rule 

\(\lim _{t \rightarrow-\infty}t  .e^t=\lim _{t \rightarrow-\infty} \frac{t}{e^{-t}}\)

\(=\lim _{t \rightarrow -\infty} \frac{1}{-e^{-t}}=\frac{1}{-e^{-(-\infty)}}=\frac{1}{-e^{\infty}}=\frac{1}{-\infty}=0\)

\(=-\lim _{t \rightarrow-\infty} te^t-1+0=-1\)

Evaluate \(\int_{-\infty}^{\infty} \frac{1}{1+x^{2}} d x\)

\(\int_{-\infty}^{\infty} f(x) d(x)=\int_{-\infty}^{a} f(x) d x+\int_{a}^{-\infty} f(x) d x\)

\(a=0\)

\(\int_{-\infty}^{\infty} \frac{1}{1+x^{2}} d x=\int_{-\infty}^{0} \frac{1}{1+x^{2}} d x+\int_{0}^{\infty} \frac{1}{1+x^{2}} d x\)

\(\int \frac{1}{1+x^{2}} d x=\tan ^{-1}(x)\)

\(\int_{-\infty}^{0} \frac{1}{1+x^{2}} d x=\lim _{t \rightarrow-\infty} \int_{t}^{0} \frac{1}{1+x^{2}} d x\)

\(=\lim _{t \rightarrow-\infty} \tan ^{-1}\left.(x)\right|_{t}^{0}=\lim _{t \rightarrow-\infty}\left(\tan ^{-1}(0)-\tan ^{-1}(t)\right)\)

\(\tan ^{-1}(-\infty)=\frac{-\pi}{2}\)

\(\int_{-\infty}^{0} \frac{1}{1+x^{2}} d x=\tan ^{-1}(0)-\tan ^{-1}(-\infty)=0-\left(-\frac{\pi}{2}\right)=\frac{\pi}{2}\)

\(\int_{0}^{\infty} \frac{1}{1+x^{2}} d x=\lim _{t \rightarrow \infty} \int_{0}^{t} \frac{d x}{1+x^{2}}\)

\(=\lim _{t \rightarrow \infty} \tan ^{-1}(x) |_{0}^{t}=\lim _{t \rightarrow \infty}\left[\tan ^{-1}(t)-\tan ^{-1}(0)\right]\)

\(=\lim _{t \rightarrow \infty}\left[\tan ^{-1}(t)\right]=\frac{\pi}{2}\)

\(\tan ^{-1}(\infty)=\frac{\pi}{2}\)

\(\int_{-\infty}^{\infty} \frac{1}{1+x^{2}} d x=\int_{-\infty}^{0} \frac{1}{1+x^{2}} d x+\int_{0}^{\infty} \frac{1}{1+x^{2}} d x\)

\(=\frac{\pi}{2}+\frac{\pi}{2}=\frac{2 \pi}{2}=\pi\)

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