Integration by Parts

Example 2.

1. For \(a,b\in\mathbb{R}\), determine \[\int_a^bx\exp(x)dx.\] We use integration by parts with \(f'(x)=f(x)=\exp(x)\) and \(g(x)=x\). Then \[\begin{aligned} \int_a^b\exp(x)xdx=&\left.x\exp(x)\right|_{x=a}^{x=b}-\int_a^b\exp(x)dx\\ =&\left.x\exp(x)\right|_{x=a}^{x=b}-\left.\exp(x)\right|_{x=a}^{x=b}\\=&\left.(x-1)\exp(x)\right|_{x=a}^{x=b}. \end{aligned}\] It is very important to note that an unlucky choice of \(f\) and \(g\) may be misleading. For instance, if we choose \(f'(x)=x\) and \(g(x)=\exp(x)\). Then integration by parts gives \[\int_a^bx\exp(x)dx=\left.\frac12x^2\exp(x)\right|_{x=a}^{x=b}-\int_a^b\frac12x^{2}\exp(x)dx.\] This formula is mathematically correct, but it does not lead to the explicit determination of the integral.

2. Integrate \(\sin(x)\cos(x)\): \[\begin{aligned} &\int_a^b\underbrace{\sin(x)}_{f'(x)=-\cos'(x)}\underbrace{\cos(x)}_{=g(x)}dx\\=&\left.-\cos^2(x)\right|_{x=a}^{x=b}-\int_a^b(-\cos(x))(-\sin(x))dx\\ =&\left.-\cos^2(x)\right|_{x=a}^{x=b}-\int_a^b\cos(x)\sin(x)dx. \end{aligned}\] Solving this equation for \(\int_a^b\sin(x)\cos(x)dx\), we obtain \[\int_a^b\sin(x)\cos(x)dx=\left.-\frac12\cos^2(x)\right|_{x=a}^{x=b}.\]

3. For \(a,b\in(0,\infty)\) and \(\lambda\in\mathbb{R}\backslash\{-1\}\), determine the integral \[\int_a^bx^\lambda\log(x)dx.\] Defining \(f(x)=\frac1{\lambda+1}x^{\lambda+1}\), \(g(x)=\log(x)\), integration by parts leads to \[\begin{aligned} &\int_a^bx^\lambda\log(x)dx\\ =&\left.\frac1{\lambda+1}x^{\lambda+1}\log(x)\right|_{x=a}^{x=b}-\int_a^b\frac1{\lambda+1}x^{\lambda+1}\frac1xdx\\ =&\left.\frac1{\lambda+1}x^{\lambda+1}\log(x)\right|_{x=a}^{x=b}-\int_a^b\frac1{\lambda+1}x^{\lambda}dx\\ =&\left.\frac1{\lambda+1}x^{\lambda+1}\log(x)\right|_{x=a}^{x=b}-\left.\frac1{(\lambda+1)^2}x^{\lambda+1}\right|_{x=a}^{x=b}\\ =&\left.\frac1{\lambda+1}x^{\lambda+1}\left(\log(x)-\frac1{\lambda+1}\right)\right|_{x=a}^{x=b}. \end{aligned}\]

4. For \(a,b\in(0,\infty)\), determine the integral \[\int_a^b\frac{\log(x)}xdx.\] Defining \(f(x)=g(x)=\log(x)\), integration by parts leads to \[\int_a^b\frac{\log(x)}xdx =\left.\log^2(x)\right|_{x=a}^{x=b}-\int_a^b\frac{\log(x)}xdx.\] Solving this for \(\int_a^b\frac{\log(x)}xdx\), we obtain \[\int_a^b\frac{\log(x)}xdx =\frac12\left.\log^2(x)\right|_{x=a}^{x=b}.\]