Calculus – Integration – Integration by parts

Using the product rule we have

\[\frac{d}{dx}\left\{f(x)g(x)\right\}=f(x)g'(x)+g(x)f'(x).\]

Hence we have

\[f(x)g(x)=\int f(x)g'(x)\,dx+\int g(x)f'(x)\,dx.\]

Now we use \(g'(x)\,dx=dg(x)\) and \(f'(x)\,dx=df(x)\) to obtain:

Theorem:

\[\int f(x)\,dg(x)=f(x)g(x)-\int g(x)\,df(x).\]

Theorem:

\[\int_a^b f(x)\,dg(x)=\bigg[f(x)g(x)\bigg]_a^b-\int_a^b g(x)\,df(x).\]

Serret's integral

We have seen that Serret's integral equals \(\displaystyle\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx=\frac{1}{8}\pi\ln(2)\).

Using integration by parts we obtain

\begin{align*} \int_0^1\frac{\ln(1+x)}{1+x^2}\,dx&=\int_0^1\ln(1+x)\,d\arctan(x)=\ln(1+x)\arctan(x)\bigg|_0^1-\int_0^1\arctan(x)\,d\ln(1+x)\\[2.5mm] &=\frac{1}{4}\pi\ln(2)-\int_0^1\frac{\arctan(x)}{1+x}\,dx. \end{align*}

This implies that

\[\int_0^1\frac{\arctan(x)}{1+x}\,dx=\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx=\frac{1}{8}\pi\ln(2).\]

Note that Serret's integral is a special case of \(I(\alpha):=\displaystyle\int_0^{\alpha}\frac{\ln(1+\alpha x)}{1+x^2}\,dx\).

Differentiation with respect to \(\alpha\) leads to \(I'(\alpha)=\displaystyle\frac{\ln(1+\alpha^2)}{1+\alpha^2}+\int_0^{\alpha}\frac{x}{(1+x^2)(1+\alpha x)}\,dx\). Now we use partial fractions to find that

\begin{align*} \int_0^{\alpha}\frac{x}{(1+x^2)(1+\alpha x)}\,dx&=\frac{1}{1+\alpha^2}\int_0^{\alpha}\left(\frac{\alpha+x}{1+x^2}-\frac{\alpha}{1+\alpha x}\right)\,dx =\frac{1}{1+\alpha^2}\bigg[\alpha\arctan(x)+\frac{1}{2}\ln(1+x)-\ln(1+\alpha x)\bigg]_0^{\alpha}\\[2.5mm] &=\frac{1}{1+\alpha^2}\left(\alpha\arctan(\alpha)+\frac{1}{2}\ln(1+\alpha^2)-\ln(1+\alpha^2)\right) =\frac{\alpha}{1+\alpha^2}\arctan(\alpha)-\frac{\ln(1+\alpha^2)}{2(1+\alpha^2)}. \end{align*}

Hence we have: \(I'(\alpha)=\displaystyle\frac{\alpha}{1+\alpha^2}\arctan(\alpha)+\frac{\ln(1+\alpha^2)}{2(1+\alpha^2)}\). Since \(I(0)=0\), this implies that

\begin{align*} I(\alpha)&=\int\frac{\alpha}{1+\alpha^2}\arctan(\alpha)\,d\alpha+\int\frac{\ln(1+\alpha^2)}{2(1+\alpha^2)}\,d\alpha =\frac{1}{2}\int\arctan(\alpha)\,d\ln(1+\alpha^2)+\int\frac{\ln(1+\alpha^2)}{2(1+\alpha^2)}\,d\alpha\\[2.5mm] &=\frac{1}{2}\arctan(\alpha)\ln(1+\alpha^2)-\frac{1}{2}\int\ln(1+\alpha^2)\,d\arctan(\alpha)+\frac{1}{2}\int\frac{\ln(1+\alpha^2)}{1+\alpha^2}\,d\alpha\\[2.5mm] &=\frac{1}{2}\arctan(\alpha)\ln(1+\alpha^2)-\frac{1}{2}\int\frac{\ln(1+\alpha^2)}{1+\alpha^2}\,d\alpha+\frac{1}{2}\int\frac{\ln(1+\alpha^2)}{1+\alpha^2}\,d\alpha\\[2.5mm] &=\frac{1}{2}\arctan(\alpha)\ln(1+\alpha^2)+C\quad\Longrightarrow\quad I(\alpha)=\frac{1}{2}\arctan(\alpha)\ln(1+\alpha^2). \end{align*}

Hence we have: \(\displaystyle\int_0^{\alpha}\frac{\ln(1+\alpha x)}{1+x^2}\,dx=\frac{1}{2}\arctan(\alpha)\ln(1+\alpha^2)\).

The special case \(\alpha=1\) now reads

\[\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx=\frac{1}{2}\arctan(1)\ln(2)=\frac{1}{2}\cdot\frac{1}{4}\pi\cdot\ln(2)=\frac{1}{8}\pi\ln(2).\]

Similarly, the integral \(\displaystyle\int_0^1\frac{\arctan(x)}{1+x}\,dx\) is a special case of

\begin{align*} \alpha\int_0^{\alpha}\frac{\arctan(x)}{1+\alpha x}\,dx&=\int_0^{\alpha}\arctan(x)\,d\ln(1+\alpha x) =\arctan(x)\ln(1+\alpha x)\bigg|_0^{\alpha}-\int_0^{\alpha}\ln(1+\alpha x)\,d\arctan(x)\\[2.5mm] &=\arctan(\alpha)\ln(1+\alpha^2)-\int_0^{\alpha}\frac{\ln(1+\alpha x)}{1+x^2}\,dx. \end{align*}

This implies that: \(\displaystyle\alpha\int_0^{\alpha}\frac{\arctan(x)}{1+\alpha x}\,dx=\frac{1}{2}\arctan(\alpha)\ln(1+\alpha^2) =\int_0^{\alpha}\frac{\ln(1+\alpha x)}{1+x^2}\,dx\).

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Last modified on March 8, 2021