A crucial tool in the proof of Dirichlet's theorem for convergence of Fourier series and other statements involving trigonometric integrals, is the following lemma.
If #f# is piecewise continuous on some closed interval #\ivcc{a}{b}# with #a,b \in \mathbb{R}# and #a\lt b#, then
\[\begin{array}{rcl}\displaystyle \lim_{\alpha \rightarrow \infty}{ \int_{a}^{b}} f(t)\sin(\alpha t) \dd t&=&0 \\\displaystyle \lim_{\alpha \rightarrow \infty}{ \int_{a}^{b}} f(t)\cos(\alpha t) \dd t &=&0 \end{array}\]
We prove the first equality in the special case where #f# is piecewise smooth. This proof is short. It will be used in the proof for the general case. We can assume #f# and #f'# to be continuous on \(\ivcc{a}{b}\), since the piecewise continuous case will then follow by applying the argument for the smooth case a finite number of times, once for each interval between jumps of discontinuities on which #f# is smooth.
Integration by parts gives us, for every real number #\alpha\gt0#,
\[\begin{array}{rcl} \displaystyle \int_{a}^{b} f(t) \sin(\alpha t) \dd t &=& \displaystyle \left[-\,\frac{f(t)\cos(\alpha t)}{\alpha}\right]_a^b+\frac{1}{\alpha}\int_{a}^{b} f'(t) \cos(\alpha t) \dd t\\&=& \displaystyle \frac{f(a)\cos(\alpha a)-f(b)\cos(\alpha b)}{\alpha}+\frac{1}{\alpha}\int_{a}^{b} f'(t) \cos(\alpha t) \dd t\end{array}\]
According to the Min-Max theorem, the functions #f# and #f'# have global minima and maxima on #\ivcc{a}{b}#. In particular, there is a positive number #M# such that #|f(t)|\le M# and #|f'(t)|\le M# for all #t\in \ivcc{a}{b}#. As a consequence, using Estimates of integrals, we find
\[\begin{array}{rcl} \displaystyle\left| \int_{a}^{b}{ f(t) \sin(\alpha t)} \dd t \right|&\le& \displaystyle \frac{\left|f(a)\cos(\alpha a)-f(b)\cos(\alpha b)\right|}{\alpha}+\frac{1}{\alpha}\left|\int_{a}^{b}{ f'(t) \cos(\alpha t)} \dd t \right| \\ &\le& \displaystyle \frac{2M}{\alpha}+\frac{1}{\alpha}\int_{a}^{b}\left| f'(t) \right|\dd t \\ &\le& \displaystyle \frac{2M}{\alpha}+\frac{M\cdot |b-a|}{\alpha} \\ \displaystyle&=& \displaystyle \frac{M\cdot(2+|b-a|)}{\alpha} \end{array}\]
We conclude that
\[\lim_{\alpha \rightarrow \infty}{\left|\int_{a}^{b}f(t)\sin(\alpha t) \dd t\right|} =0\]
from which it immediately follows that \( \lim_{\alpha \rightarrow \infty} {\int_{a}^{b} f(t)\sin(\alpha t) }\dd t=0\). This proves the first equality.
The proof of the second equality is omitted since it is very similar to the above.
As in the special case, and for the same reason, we only focus on the first limit. Let #\varepsilon\gt0# (an arbitrary small positive number). We need to show that there is a number #A# such that, for all #\alpha\gt A#, we have \[\left|{ \int_{a}^{b}} f(t)\sin(\alpha t) \dd t\right|\lt\varepsilon\]
The fact that #f# is piecewise continuous on #\ivcc{a}{b}# implies that the Riemann integral of #f# on #\ivcc{a}{b}# exists. In particular, there is a partitioning \[a = x_0,x_1,\ldots,x_{n-1},x_n=b\] of the interval with #x_i\lt x_{i+1}# for #i=0,\ldots,n-1# such that, with #m_i# the minimum of #f# on #\ivcc{x_i}{x_{i+1}}# and #M_i# the maximum of #f# on #\ivcc{x_i}{x_{i+1}}#, we have
\[\sum_{i=0}^{n-1}(x_{i+1}-x_i)(M_i-m_i)\lt \frac{\varepsilon}{2}\]The choice \( \frac{\varepsilon}{2}\) instead of \(\varepsilon\) is allowed because we start off with an arbitrary positive number, which we can choosen to be \( \frac{\varepsilon}{2}\). The actual integral of #f# on #\ivcc{a}{b}# lies in between the sums involving the lower bounds #m_i# and the upper bounds #M_i#. Therefore,
\[\left|\int_a^b f(x)\dd x- \sum_{i=0}^{n-1}(x_{i+1}-x_i)\cdot m_i\right|\lt \frac{\varepsilon}{2}\]
The sum \(\sum_{i=0}^{n-1}(x_{i+1}-x_i)\cdot m_i\) can be viewed as the integral #\int_a^b m(x)\dd x# of the piecewise constant function #m# with function rule
\[m(x) = m_i\text{ if } x_i\le x\le x_{i+1} \text{ for } i=0,\ldots,n-1\]Bearing in mind that #f(x)\ge m(x)# for all #x# in #\ivcc{a}{b}#, we can rewrite the above inequality in terms of the function #m#:
\[\int_a^b\left( f(x)- m(x)\right)\dd x\lt \frac{\varepsilon}{2}\]
Thus the integral of #f# can be approximated by the integral of the piecewise constant function #m#.
By the proof of the special case, the theorem holds for #m#. This means that there is a number #A# such that, for all #\alpha\gt A#, we have
\[\left|\int_a^b m(x)\sin(\alpha x) \dd x\right|\lt\frac{\varepsilon}{2}\]
As a consequence, by Estimates of integrals, we have the following chain of (in)equalities for all #\alpha\gt A#.
\[\begin{array}{rcl}\displaystyle \left|\int_a^b f(x)\sin(\alpha x)\dd x\right| &=&\displaystyle\left| \int_a^b \left(f(x)-m(x)\right) \sin(\alpha x) \dd x+ \int_a^b m(x)\sin(\alpha x) \dd x\right|\\&&\phantom{xxx}\color{blue}{\text{linearity of the integral}}\\&\le&\displaystyle\left| \int_a^b \left(f(x)-m(x)\right) \sin(\alpha x) \dd x\right|+ \left|\int_a^b m(x)\sin(\alpha x) \dd x\right| \\&&\phantom{xxx}\color{blue}{|y+z|\le |y|+|z|}\\&\le&\displaystyle \int_a^b \left|f(x)-m(x)\right| \dd x+ \left|\int_a^b m(x)\sin(\alpha x) \dd x\right| \\&&\phantom{xxx}\color{blue}{\left|\int_a^b g(x)\dd x\right|\le\int_a^b\left| g(x)\right|\dd x\text{ and }|\sin(u)| \le 1}\\&\lt&\displaystyle\frac{\varepsilon}{2}+ \frac{\varepsilon}{2}\\&&\phantom{xxx}\color{blue}{\text{two inequalities derived above}}\\&=&\varepsilon \end{array}\]
This ends the proof of the theorem.
We recall that a function on an interval is called Riemann integrable if its Riemann integral on that interval exists. All piecewise continuous functions are Riemann integrable.
The theorem is also valid for Riemann integrable functions. There exist functions that are Riemann integrable but not continuous at an infinite number of points, but we will not be concerned with such functions. Therefore, the theorem is stated only for piecewise continuous functions.
Use the Riemann-Lebesgue lemma to determine the value of the integral
#\displaystyle \lim_{\alpha \rightarrow \infty}\int_0^{\pi} t \cos(\alpha t + 16) \, \dd t #
Type "none" if the lemma cannot be applied.
#\displaystyle \lim_{\alpha \rightarrow \infty}\int_0^{\pi} t \cos(\alpha t + 16) \, \dd t # = #0#
By use of the addition formula for the cosine #\cos(x+y)=\cos(x)\cos(y) - \sin(x)\sin(y)#, the integral can be written as a linear combination of two integrals of the type to which the lemma applies. \[\begin{array}{rcl} \int_0^{\pi} t \cos(\alpha t + 16) \dd t &=&\displaystyle \cos(16)\int_0^{\pi} t\cos(\alpha t) \dd t - \sin(16)\int_0^{\pi} t\sin(\alpha t)\, \dd t\\&&\phantom{xxx}\color{blue}{\cos(x+y)=\cos(x)\cos(y) - \sin(x)\sin(y)} \\ &=& 0 -0\\ &&\phantom{xx}\color{blue}{\text{Riemann-Lebesgue lemma applied twice}}\\ &=& 0\end{array}\]
Therefore, #\displaystyle \lim_{\alpha \rightarrow \infty}\int_0^{\pi} t \cos(\alpha t + 16) \dd t=0 #.
The Riemann-Lebesgue lemma
