Calculus – Multiple integrals – More applications 2

During the proof of the Gaussian integral we already saw that improper double integrals can be treated in the same way as single integrals. Here we will consider some examples of single integrals, which can be evaluated more easily by first converting them into multiple integrals.

The Dirichlet integral

Using the Feynman method we have already shown that \(\displaystyle\int_0^{\infty}\frac{\sin(x)}{x}\,dx=\tfrac{1}{2}\pi\). This can also be done through double integrals as follows: note that \(\displaystyle\frac{\sin(x)}{x}=\int_0^1\cos(xt)\,dt\) for \(x>0\). Then we have using integration by parts:

\begin{align*} \int_0^{\infty}e^{-\alpha x}\frac{\sin(x)}{x}\,dx&=\int_0^{\infty}e^{-\alpha x}\int_0^1\cos(xt)\,dt\,dx=\int_0^1\int_0^{\infty}e^{-\alpha x}\cos(xt)\,dx\,dt\\[2.5mm] &=\int_0^1\frac{\alpha}{\alpha^2+t^2}\,dt=\arctan\left(\frac{t}{\alpha}\right)\bigg|_0^1=\arctan\left(\frac{1}{\alpha}\right),\quad\alpha>0. \end{align*}

Finally we have: \(\displaystyle\int_0^{\infty}\frac{\sin(x)}{x}\,dx=\lim\limits_{\alpha\downarrow0}\int_0^{\infty}e^{-\alpha x}\frac{\sin(x)}{x}\,dx =\lim\limits_{\alpha\downarrow0}\arctan\left(\frac{1}{\alpha}\right)=\tfrac{1}{2}\pi\).

Similarly we have \(\displaystyle\int_0^1\sin(xt)\,dt=\frac{1-\cos(x)}{x}\) and therefore we have for \(\alpha>0\)

\begin{align*} \int_0^{\infty}e^{-\alpha x}\frac{1-\cos(x)}{x}\,dx&=\int_0^{\infty}e^{-\alpha x}\int_0^1\sin(xt)\,dt\,dx=\int_0^1\int_0^{\infty}e^{-\alpha x}\sin(xt)\,dx\,dt\\[2.5mm] &=\int_0^1\frac{t}{\alpha^2+t^2}\,dt=\frac{1}{2}\ln(\alpha^2+t^2)\bigg|_0^1=\frac{1}{2}\ln(\alpha^2+1)-\ln(\alpha). \end{align*}

For instance, this implies that \(\displaystyle\int_0^{\infty}\frac{e^{-x}(1-\cos(x))}{x}\,dx=\tfrac{1}{2}\ln(2)\).

Note that \(\displaystyle\frac{\partial}{\partial t}\left(\frac{\arctan(xt)}{x}\right)=\frac{1}{1+x^2t^2}\). Then we have for \(0<\alpha<\beta\):

\begin{align*} \int_0^{\infty}\frac{\arctan(\beta x)-\arctan(\alpha x)}{x}\,dx&=\int_0^{\infty}\int_{\alpha}^{\beta}\frac{1}{1+x^2t^2}\,dt\,dx =\int_{\alpha}^{\beta}\int_0^{\infty}\frac{1}{1+x^2t^2}\,dx\,dt\\[2.5mm] &=\int_{\alpha}^{\beta}\left[\frac{\arctan(xt)}{t}\right]_{x=0}^{\infty}\,dt=\frac{1}{2}\pi\int_{\alpha}^{\beta}\frac{dt}{t} =\frac{1}{2}\pi\left(\ln(\beta)-\ln(\alpha)\right). \end{align*}

Similarly we obtain for \(0<\alpha<\beta\):

\begin{align*} \int_0^{\infty}\frac{e^{-\alpha x}-e^{-\beta x}}{x}\,dx&=\int_0^{\infty}\int_{\alpha}^{\beta}e^{-xt}\,dt\,dx=\int_{\alpha}^{\beta}\int_0^{\infty}e^{-xt}\,dx\,dt\\[2.5mm] &=\int_{\alpha}^{\beta}\left[-\frac{e^{-xt}}{t}\right]_{x=0}^{\infty}=\int_{\alpha}^{\beta}\frac{dt}{t}=\ln(\beta)-\ln(\alpha). \end{align*}

Using integration by parts we obtain for \(0<\alpha<\beta\)

\begin{align*} \int_0^{\infty}\frac{(e^{-\alpha x}-e^{-\beta x})\sin(x)}{x}\,dx&=\int_0^{\infty}\sin(x)\int_{\alpha}^{\beta}e^{-xt}\,dt\,dx =\int_{\alpha}^{\beta}\int_0^{\infty}e^{-xt}\sin(x)\,dx\,dt\\[2.5mm] &=\int_{\alpha}^{\beta}\frac{dt}{1+t^2}=\arctan(\beta)-\arctan(\alpha) \end{align*}

and

\begin{align*} \int_0^{\infty}\frac{(e^{-\alpha x}-e^{-\beta x})\cos(x)}{x}\,dx&=\int_0^{\infty}\cos(x)\int_{\alpha}^{\beta}e^{-xt}\,dt\,dx =\int_{\alpha}^{\beta}\int_0^{\infty}e^{-xt}\cos(x)\,dx\,dt\\[2.5mm] &=\int_{\alpha}^{\beta}\frac{t}{1+t^2}\,dt=\frac{1}{2}\ln(1+\beta^2)-\frac{1}{2}\ln(1+\alpha^2). \end{align*}

An example of a triple integral: for \(0<\alpha<\beta\) we have

\begin{align*} \int_0^{\infty}\frac{(1-e^{-\alpha x})(1-e^{-\beta x})}{x^2}\,dx&=\int_0^{\infty}\left(\int_0^{\alpha}e^{-xy}\,dy\right)\left(\int_0^{\beta}e^{-xz}\,dz\right)\,dx =\int_0^{\alpha}\int_0^{\beta}\left(\int_0^{\infty}e^{-x(y+z)}\,dx\right)\,dz\,dy\\[2.5mm] &=\int_0^{\alpha}\int_0^{\beta}\frac{1}{y+z}\,dz\,dy=\int_0^{\alpha}\left(\ln(y+\beta)-\ln(y)\right)\,dy\\[2.5mm] &=\bigg[(y+\beta)\ln(y+\beta)-y\ln(y)\bigg]_0^{\alpha}=(\alpha+\beta)\ln(\alpha+\beta)-\alpha\ln(\alpha)-\beta\ln(\beta). \end{align*}

Finally, we consider the integral \(\displaystyle\int_0^{\frac{1}{2}\pi}\ln\left(1+\alpha\cos^2(x)\right)\,dx\) for \(\alpha>0\). Since \(\displaystyle\frac{\partial}{\partial y}\ln\left(1+y\cos^2(x)\right)=\frac{\cos^2(x)}{1+y\cos^2(x)}\) we find for \(\alpha>0\)

\[\int_0^{\frac{1}{2}\pi}\ln\left(1+\alpha\cos^2(x)\right)\,dx=\int_0^{\frac{1}{2}\pi}\int_0^{\alpha}\frac{\cos^2(x)}{1+y\cos^2(x)}\,dy\,dx =\int_0^{\alpha}\int_0^{\frac{1}{2}\pi}\frac{\cos^2(x)}{1+y\cos^2(x)}\,dx\,dy.\]

Now we use that \(\displaystyle\frac{1}{\cos^2(x)}=\frac{\sin^2(x)+\cos^2(x)}{\cos^2(x)}=1+\tan^2(x)\) and find that

\[\int_0^{\frac{1}{2}\pi}\frac{\cos^2(x)}{1+y\cos^2(x)}\,dx=\int_0^{\frac{1}{2}\pi}\frac{1}{1+y+\tan^2(x)}\,dx=\int_0^{\infty}\frac{1}{1+y+t^2}\,d\arctan(t) =\int_0^{\infty}\frac{dt}{(1+y+t^2)(1+t^2)}.\]

Using partial fractions we now obtain for \(y>0\)

\begin{align*} \int_0^{\infty}\frac{dt}{(1+y+t^2)(1+t^2)}&=\frac{1}{y}\int_0^{\infty}\left(\frac{1}{1+t^2}-\frac{1}{1+y+t^2}\right)\,dt\\[2.5mm] &=\frac{1}{y}\left[\arctan(t)-\frac{1}{\sqrt{1+y}}\arctan\left(\frac{t}{\sqrt{1+y}}\right)\right]_{t=0}^{\infty} =\frac{1}{2}\pi\cdot\frac{1}{y}\left(1-\frac{1}{\sqrt{1+y}}\right). \end{align*}

Now we use the substitution \(\sqrt{1+y}=t\) or \(y=t^2-1\) and obtain that

\[\int_0^{\alpha}\frac{1}{y}\left(1-\frac{1}{\sqrt{1+y}}\right)\,dy=\int_1^{\sqrt{1+\alpha}}\frac{1}{t^2-1}\left(1-\frac{1}{t}\right)2t\,dt =2\int_0^{\sqrt{1+\alpha}}\frac{dt}{1+t}=2\ln\left(\frac{1+\sqrt{1+\alpha}}{2}\right).\]

Hence: \(\displaystyle\int_0^{\frac{1}{2}\pi}\ln\left(1+\alpha\cos^2(x)\right)\,dx=\pi\ln\left(\frac{1+\sqrt{1+\alpha}}{2}\right)\) for \(\alpha>0\).

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Last modified on April 20, 2024