Limit of integrals (15)

Problem. Show that

i) \displaystyle u-\frac{u^3}{6} \le \sin u \le u for u \ge 0 and \displaystyle \cos u \ge 1- \frac{u^2}{2} for all real numbers u

ii) \displaystyle \lim_{x\to+\infty} \int_0^x \sin\left(\frac{1}{x+t}\right) dt=\ln 2 and \displaystyle \lim_{x\to+\infty} \int_0^x \cos\left(\frac{1}{x+t}\right) dt=+\infty.

Solution. i) Let

\displaystyle f(u):=u-\sin u, \ \ g(u):=\sin u - u+\frac{u^3}{6}, \ \ h(u):=\cos u - 1 + \frac{u^2}{2}.

The function f is increasing on the real line because f'(u)=1-\cos u \ge 0. So f(u) \ge f(0)=0 for u \ge 0.
We have h'(u)=u-\sin u=f(u). Therefore h is increasing for u \ge 0 and so h(u) \ge h(0)=0 for u \ge 0. Since h is an even function, h(u) \ge 0 for all real numbers u.
We have \displaystyle g'(u)=\cos u - 1 + \frac{u^2}{2}=h(u). So g is increasing for all real numbers u and hence g(u) \ge g(0)=0 for u \ge 0.

ii) By i), \displaystyle \frac{1}{x+t} - \frac{1}{6(x+t)^3} \le \sin\left(\frac{1}{x+t}\right) \le \frac{1}{x+t} for all x,t > 0. So if x > 0, then

\displaystyle \begin{aligned} \ln 2 - \frac{1}{16x^2}=\int_0^x \frac{dt}{x+t} -\int_0^x \frac{dt}{6(x+t)^3} \le \int_0^x \sin\left(\frac{1}{x+t}\right) dt \le \int_0^x \frac{dt}{x+t}=\ln 2\end{aligned}

and the result follows from the squeeze theorem.
Also, again by i), \displaystyle \cos\left(\frac{1}{x+t}\right) \ge 1 - \frac{1}{2(x+t)^2} for all x,t > 0. So if x > 0, then

\displaystyle \int_0^x \cos\left(\frac{1}{x+t}\right) dt \ge \int_0^xdt - \int_0^x \frac{dt}{2(x+t)^2}=x-\frac{1}{4x}

and the result follows. \Box

List of Maclaurin series for some functions

\displaystyle(x+1)^a=\sum_{n=0}^{\infty}\binom{a}{n}x^n, \ \ \ \ \ a \in \mathbb{R}, \ \ x \in (-1,1)

Comment. See here!


\displaystyle \frac{1}{x+1}=\sum_{n=0}^{\infty}(-1)^nx^n, \ \ \ \ \ x \in (-1,1)


\displaystyle \begin{aligned} \frac{1}{(x+1)^m}=\sum_{n=0}^{\infty}(-1)^n \binom{n+m-1}{n}x^n, \ \ \ \ \ m \ \text{positive integer}, \ x \in (-1,1)\end{aligned}


\displaystyle \sqrt{x+1}=\sum_{n=0}^{\infty}\frac{(-1)^{n-1}\binom{2n}{n}}{(2n-1)4^n}x^n, \ \ \ \ \ x \in (-1,1)


\displaystyle \frac{1}{\sqrt{x+1}}=\sum_{n=0}^{\infty}\frac{(-1)^n\binom{2n}{n}}{4^n}x^n, \ \ \ \ \ x \in (-1,1)


\displaystyle \begin{aligned} \frac{1}{(x+\cot \theta)^2+1}=\sum_{n=0}^{\infty}(-1)^n \sin((n+1)\theta)(\sin \theta)^{n+1}x^n, \ \ \ \ \ \theta \in (0, \pi), \ x \in (-\csc \theta, \csc \theta) \end{aligned}

Comment. See here!


\displaystyle \frac{1}{1-x-x^2}=\sum_{n=0}^{\infty}F_nx^n, \ \ \ \ \ x \in \left(\frac{1-\sqrt{5}}{2}, \frac{\sqrt{5}-1}{2}\right)

Comment. See here and here!


\displaystyle \ln(x+1)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}x^n, \ \ \ \ \ x \in (-1,1]


\displaystyle (\ln(x+1))^2=2\sum_{n=2}^{\infty} \left(\frac{(-1)^n}{n}\sum_{k=1}^{n-1} \frac{1}{k}\right)x^n, \ \ \ \ \  x \in (-1,1]

Comment. It follows from \displaystyle (\ln(x+1))^2=2\int_0^x \frac{\ln(t+1)}{t+1} \ dt.


\displaystyle e^{e^x} = \sum_{n=0}^{\infty} \frac{b_ne}{n!}x^n, \ \ \ \ \ x \in \mathbb{R}

Comment. See here! (Example 2)


\displaystyle \frac{x}{e^x-1}=\sum_{n=0}^{\infty} \frac{B_n}{n!}x^n, \ \ \ \ \ x \in (-2\pi,2\pi)

Comment. See here!


\displaystyle \sinh x = \frac{e^x-e^{-x}}{2}=\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}, \ \ \ \ \ x \in \mathbb{R}


\displaystyle \cosh x = \frac{e^x+e^{-x}}{2}=\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}, \ \ \ \ \ x \in \mathbb{R}


\displaystyle \tanh x = \sum_{n=1}^{\infty} \frac{4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-1}, \ \ \ \ \ x \in \left(\frac{-\pi}{2}, \frac{\pi}{2}\right)

Comment. The proof is similar to the one given for the Maclurin series of \tan x. Note that \displaystyle \coth x has no Maclaurin series expansion because it is not even defined at x=0.


\displaystyle \sinh^{-1}x=\sum_{n=0}^{\infty} \frac{(-1)^n\binom{2n}{n}}{(2n+1)4^n}x^{2n+1}, \ \ \ \ \ x \in (-1,1)


\displaystyle \tanh^{-1} x=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)=\sum_{n=1}^{\infty}\frac{x^{2n-1}}{2n-1}, \ \ \ \ \ x \in (-1,1)


\displaystyle \sin x = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!}x^{2n+1}, \ \ \ \ \ x \in \mathbb{R}


\displaystyle \cos x = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!}x^{2n}, \ \ \ \ \ x \in \mathbb{R}


\displaystyle \tan x = \sum_{n=1}^{\infty} (-1)^{n-1}\frac{4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-1}, \ \ \ \ \ x \in \left(\frac{-\pi}{2}, \frac{\pi}{2}\right)

Comment.  See here! Note that \displaystyle \cot x has no Maclaurin series expansion because it is not even defined at x=0.


\displaystyle \sin^{-1}x = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{(2n+1)4^n}x^{2n+1}, \ \ \ \ \  x \in [-1,1]

Comment. See here! Note that the Maclaurin series for \displaystyle \cos^{-1}x now follows from the identity \displaystyle \cos^{-1}x = \frac{\pi}{2}-\sin^{-1}x.


\displaystyle (\sin^{-1}x)^2=\sum_{n=1}^{\infty} \frac{2^{2n-1}}{n^2\binom{2n}{n}}x^{2n}, \ \ \ \ \ x \in [-1,1]

Comment: see here!


\displaystyle \tan^{-1}x = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}, \ \ \ \ \ x \in [-1,1]

Comment. Note that the Maclaurin series for \displaystyle \cot^{-1}x now follows from the identity \displaystyle \cot^{-1}x = \frac{\pi}{2}-\tan^{-1}x.


\displaystyle (\tan^{-1}x)^2=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n-1}}{n}\sum_{k=1}^n \frac{1}{2k-1}\right)x^{2n}, \ \ \ \ \ x \in [-1,1]

Comment. It follows from \displaystyle (\tan^{-1}x)^2=2\int_0^x \frac{\tan^{-1}t}{t^2+1} \ dt.

Integral of x^k*ln(sin(x))

Notation. Throughout this post, \displaystyle I_k:=\int_0^{\frac{\pi}{2}}x^k\ln(\sin x) \ dx, where k is an integer.

We showed here that \displaystyle I_0=\int_0^{\frac{\pi}{2}} \ln(\sin x) \ dx=-\frac{\pi}{2}\ln 2, which was done quite easily. Now, let’s ask a more interesting question: given an integer k \ge 0, how can we evaluate \displaystyle I_k=\int_0^{\frac{\pi}{2}}x^k\ln(\sin x) \ dx ?

What is certain is that the method we used to evaluate I_k for k=0 is absolutely useless for k > 0. We need a more powerful tool and we have already seen that tool in this post. As I mentioned in the Remark in that post, we have

\displaystyle \ln(\sin x)= -\ln 2 -\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n}, \ \ \ \ \ x \in (0,\pi) \ \ \ \ \ \ \ \ \ \ \ \ (1)

and thus

\displaystyle I_k= -\frac{\ln 2}{k+1}\left(\frac{\pi}{2}\right)^{k+1} - \sum_{n=1}^{\infty}\frac{1}{n}\int_0^{\frac{\pi}{2}} x^k\cos(2nx) \ dx. \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

Of course, in order to get (2), we integrated the infinite series \displaystyle \sum_{n=1}^{\infty} \frac{\cos(2nx)}{n} in (1) term by term. But can we do that? Yes, we can but I won’t explain the reason here.

Problem. Given integers k \ge 0, \ n \ge 1, let \displaystyle J_k:=\int_0^{\frac{\pi}{2}}x^k \cos(2nx) \ dx, Show that

i) \displaystyle J_0=0, \ \ J_1=\frac{(-1)^n-1}{4n^2} and

\displaystyle J_k=\frac{(-1)^nk}{4n^2}\left(\frac{\pi}{2}\right)^{k-1}-\frac{k(k-1)}{4n^2}J_{k-2}, \ \ k \ge 2.

ii) \displaystyle I_0=-\frac{\pi}{2}\ln 2, \ \ I_1=-\frac{\pi^2}{8}\ln 2+\frac{7}{16}\zeta(3) and

\displaystyle \begin{aligned} I_k=-\frac{\pi^{k+1}}{(k+1)2^{k+1}}\ln 2 + \frac{3k\pi^{k-1}}{2^{k+3}}\zeta(3)+\frac{k(k-1)}{4}\sum_{n=1}^{\infty}\frac{1}{n^3}J_{k-2}, \ \ k \ge 2.\end{aligned}

Solution. i) Integration by parts.

ii) It follows easily from i), (2) and the first part of Problem 1 in this post! \Box

Example 1. Evaluate \displaystyle \int_0^{\frac{\pi}{2}}x^2\ln(\sin x) \ dx and \displaystyle \int_0^{\frac{\pi}{2}}x^3\ln(\sin x) \ dx.

Solution. By the second part of the above problem,

\displaystyle I_2=-\frac{\pi^3}{24}\ln 2 + \frac{3\pi}{16}\zeta(3)+\frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n^3}J_0.

By the first part of the above problem, J_0=0 and so \displaystyle I_2=-\frac{\pi^3}{24}\ln 2 + \frac{3\pi}{16}\zeta(3).
Again, by the second part of the above problem,

\displaystyle I_3= - \frac{\pi^4}{64}\ln 2 + \frac{9\pi^2}{64}\zeta(3)+\frac{3}{2}\sum_{n=1}^{\infty} \frac{1}{n^3}J_1. \ \ \ \ \ \ \ \ \ \ \ \ (3)

By the first part of the above problem, \displaystyle J_1=\frac{(-1)^n-1}{4n^2} and so by the first part of Problem 1 in this post, we have

\displaystyle \begin{aligned}\sum_{n=1}^{\infty} \frac{1}{n^3}J_1=\frac{1}{4}\left(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^5}-\sum_{n=1}^{\infty} \frac{1}{n^5}\right)=-\frac{1}{4}((1-2^{-4})\zeta(5)+\zeta(5))=-\frac{31}{64}\zeta(5).\end{aligned}

Thus, by \displaystyle (3), \ I_3= - \frac{\pi^4}{64}\ln 2 + \frac{9\pi^2}{64}\zeta(3)-\frac{93}{128}\zeta(5). \ \Box

Example 2. Evaluate \displaystyle \int_0^{\pi}x^2\ln(\sin x) \ dx.

Solution. We have

\displaystyle \begin{aligned} \int_0^{\pi}x^2\ln(\sin x) \ dx= \int_0^{\frac{\pi}{2}}x^2\ln(\sin x) \ dx + \int_{\frac{\pi}{2}}^{\pi}x^2\ln(\sin x) \ dx \\ =I_2 +\int_0^{\frac{\pi}{2}}(\pi-x)^2\ln(\sin x) \ dx \\ =\pi^2I_0-2\pi I_1+2I_2. \end{aligned}

The values of I_0,I_1 are given in the second part of the above problem and we found I_2 in the above example. \Box

The Riemann zeta function

The Riemann zeta function \zeta is defined by \displaystyle \zeta(\alpha)=\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}, where \alpha > 1 is a real number. Note that \zeta(\alpha) is well-defined because the series  \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}, is convergent for \alpha > 1. It is known that if k \ge 1 is an integer, then

\displaystyle \zeta(2k)=(-1)^{k+1} \frac{2^{2k-1}\pi^{2k}B_{2k}}{(2k)!},

where B_{2k} are Bernoulli numbers. So, for example,

\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}=\zeta(2)=\frac{2\pi^2B_2}{2!}=\frac{\pi^2}{6}, \ \ \  \sum_{n=1}^{\infty} \frac{1}{n^4}=\zeta(4)=\frac{-8\pi^4B_4}{4!}=\frac{\pi^4}{90}.

Here we saw a very elementary proof of \displaystyle \zeta(2)=\frac{\pi^2}{6}.
There is no closed form for \zeta(2k+1) but the number \zeta(3) \approx 1.202 is lucky enough to have a name; it’s called Apéry’s constant.

Problem 1. Show that

i) \displaystyle \sum_{n=1}^{\infty} \frac{1}{(2n-1)^{\alpha}}=(1-2^{-\alpha})\zeta(\alpha) and \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{\alpha}}=(1-2^{1-\alpha})\zeta(\alpha),

ii) \displaystyle 1 +2^{-\alpha} < \zeta(\alpha) < \frac{1}{1-2^{1-\alpha}},

iii) \displaystyle \lim_{\alpha\to\infty} \zeta(\alpha)=1 and \displaystyle \lim_{\alpha\to\infty} (\zeta(\alpha)-1)^{\frac{1}{\alpha}}=\frac{1}{2}.

Solution. i) We have \displaystyle \sum_{n=1}^{\infty} \frac{1}{(2n-1)^{\alpha}}=\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}-\sum_{n=1}^{\infty} \frac{1}{(2n)^{\alpha}}=\zeta(\alpha)-2^{-\alpha}\zeta(\alpha).
And for the other one

\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{\alpha}}=\sum_{n=1}^{\infty} \frac{1}{(2n-1)^{\alpha}}-\sum_{n=1}^{\infty} \frac{1}{(2n)^{\alpha}}=(1-2^{-\alpha})\zeta(\alpha)-2^{-\alpha}\zeta(\alpha).\end{aligned}

ii) It’s clear that \displaystyle \zeta(\alpha)=\sum_{n=1}^{\infty} \frac{!}{n^{\alpha}} >  1+ \frac{1}{2^{\alpha}}. We also have

\displaystyle \begin{aligned} \zeta(\alpha)=1+\sum_{n=1}^{\infty}\left(\frac{1}{(2n)^{\alpha}}+\frac{1}{(2n+1)^{\alpha}}\right) < 1+ \sum_{n=1}^{\infty} \frac{2}{(2n)^{\alpha}}=1+2^{1-n}\zeta(\alpha) \end{aligned}

and thus \displaystyle \zeta(\alpha) < \frac{1}{1-2^{1-\alpha}}.

iii) Trivial results of ii) and the squeeze theorem. \Box

Problem 2. Let m > 0 be an integer. Show that

i) \displaystyle \int_0^{\infty} \frac{x^me^{-x}}{1-e^{-x}} \ dx = m! \zeta(m+1)

ii) \displaystyle \int_0^{\infty} \frac{x^me^{-x}}{1-e^{-2x}} \ dx = m!(1-2^{-m-1})\zeta(m+1).

Solution. The solution is based on the fact that \displaystyle \int_0^{\infty}x^me^{-ax} dx = \frac{m!}{a^{m+1}} for integers m \ge 0 and real numbers a>0 (see Exercise 2 in this post!).
i) We have

\displaystyle \begin{aligned} \int_0^{\infty} \frac{x^me^{-x}}{1-e^{-x}} \ dx=\int_0^{\infty}x^me^{-x}\sum_{k=0}^{\infty}e^{-kx}dx = \sum_{k=0}^{\infty}\int_0^{\infty}x^me^{-(k+1)x}dx \\ = \sum_{k=0}^{\infty} \frac{m!}{(k+1)^{m+1}}=m!\zeta(m+1).\end{aligned}

ii) We have

\displaystyle \begin{aligned} \int_0^{\infty} \frac{x^me^{-x}}{1-e^{-2x}} \ dx=\int_0^{\infty}x^me^{-x}\sum_{k=0}^{\infty}e^{-2kx}dx = \sum_{k=0}^{\infty}\int_0^{\infty}x^me^{-(2k+1)x}dx = \sum_{k=0}^{\infty} \frac{m!}{(2k+1)^{m+1}}\end{aligned}

and the result follows from Problem 1, i). \Box

Problem 3. Show that

i) \displaystyle \sum_{k=2}^{\infty}(\zeta(k)-1)=1

ii) \displaystyle \sum_{k=2}^{\infty} \frac{\zeta(k)-1}{k}=1-\gamma, where \gamma is the Euler’s constant.

Solution. i) We have

\displaystyle \sum_{k=2}^{\infty}(\zeta(k)-1)=\sum_{k=2}^{\infty}\sum_{n=2}^{\infty} \frac{1}{n^k}=\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{1}{n^k}=\sum_{n=2}^{\infty}\frac{1}{n(n-1)}=1.

Note that \displaystyle \sum_{k=2}^{\infty}\frac{1}{n^k} is a geometric series and \displaystyle \sum_{n=2}^{\infty}\frac{1}{n(n-1)}=1 is a telescoping series.

ii) We have

\displaystyle \sum_{k=2}^{\infty} \frac{\zeta(k)-1}{k}=\sum_{k=2}^{\infty}\sum_{n=2}^{\infty} \frac{1}{kn^k}=\sum_{n=2}^{\infty}\sum_{k=2}^{\infty} \frac{1}{kn^k}=-\sum_{n=2}^{\infty}\left(\ln\left(1-\frac{1}{n}\right)+\frac{1}{n}\right). \ \ \ \ \ \ \ \ (*)

Note that the last equality in (1) follows if in the Maclaurin series expansion of \displaystyle \ln(1-x), we put \displaystyle x = \frac{1}{n}, \ n \ge 2.
Now for integer N \ge 2 we have

\displaystyle \begin{aligned} \sum_{n=2}^N\left(\ln\left(1-\frac{1}{n}\right)+\frac{1}{n}\right)=\sum_{n=2}^N\left(\ln(n-1)-\ln n\right) + \sum_{n=2}^N \frac{1}{n}=-\ln N + \sum_{n=1}^{\infty}\frac{1}{n}-1\end{aligned}

and thus

\displaystyle \begin{aligned} \sum_{n=2}^{\infty}\left(\ln\left(1-\frac{1}{n}\right)+\frac{1}{n}\right)=\lim_{N\to\infty}\sum_{n=2}^N\left(\ln\left(1-\frac{1}{n}\right)+\frac{1}{n}\right) =-1+\lim_{N\to\infty}\left(\sum_{n=1}^{\infty}\frac{1}{n}-\ln N\right) \\ =-1+\gamma.\end{aligned}

The result now follows from (*). \ \Box

Exercise 1. Evaluate \displaystyle \zeta(6) using the general formula for \zeta(2k) given at the beginning of this post and compare the result with the bounds for \zeta(6) given in Problem 1, ii).
Hint. \displaystyle B_6=\frac{1}{42} (see Exercise 3 in this post!).

Exercise 2. This exercises generalizes the results given in Problem 2. For the definition of the gamma function \Gamma(x) see this post.
Let \alpha > 0 be a real number. Show that

i) \displaystyle \int_0^{\infty} \frac{x^{\alpha}e^{-x}}{1-e^{-x}} \ dx =\Gamma(\alpha+1) \zeta(\alpha+1)

ii) \displaystyle \int_0^{\infty} \frac{x^{\alpha}e^{-x}}{1-e^{-2x}} \ dx = \Gamma(\alpha+1)(1-2^{-\alpha-1})\zeta(\alpha+1).

A useful little remark on lim (1+a_n)^n

Problem. Let \{a_n\} be a sequence and suppose that \displaystyle \lim_{n\to\infty} na_n=a \in (-\infty, +\infty]. Show that \displaystyle \lim_{n\to\infty} (1+a_n)^n=e^a.

Solution. We consider two cases.
Case 1: a=+\infty. So for any number M > 0, we have na_n > M if n is large enough. Thus \displaystyle (1+a_n)^n > \left(1+\frac{M}{n}\right)^n and therefore, because \displaystyle \lim_{n\to\infty} \left(1+\frac{M}{n}\right)^n=e^M and \displaystyle \lim_{M\to+\infty}e^M=+\infty, we have \displaystyle \lim_{n\to\infty} (1+a_n)^n=+\infty.

Case 2: a \in (-\infty,+\infty). So given \epsilon > 0, there exists N > 0 such that \displaystyle |na_n - a| < \epsilon whenever n \ge N. Thus \displaystyle 1+\frac{a-\epsilon}{n} < 1+a_n < 1+\frac{a+\epsilon}{n}. Note that since a is finite, \displaystyle 1+\frac{a-\epsilon}{n} \ge 0 if n is large enough and so

\displaystyle \left(1+\frac{a-\epsilon}{n}\right)^n < (1+a_n)^n < \left(1+\frac{a+\epsilon}{n}\right)^n.

Thus \displaystyle \lim_{n\to\infty}(1+a_n)^n =e^a because \displaystyle \lim_{n\to\infty} \left(1+\frac{a \pm \epsilon}{n}\right)^n=e^{a \pm \epsilon} and \displaystyle \lim_{\epsilon\to0^{+}} e^{a \pm \epsilon}=e^a. \ \Box

Example 1. Given an integer k \ge 0, we have \displaystyle \lim_{n\to\infty}n\sum_{i=0}^k \frac{1}{n+i}=k+1 and thus, by the above problem, \displaystyle \lim_{n\to\infty} \left(1+\sum_{i=0}^k\frac{1}{n+i}\right)^n=e^{k+1}.

Example 2. Show that \displaystyle \lim_{n\to\infty} \left(1+\left|\ln 2 + \sum_{k=1}^n \frac{(-1)^k}{k} \right|\right)^n=\sqrt{e}.

Solution. We have \displaystyle \begin{aligned}\frac{1}{1+x}=\sum_{k=0}^{n-1} (-1)^k x^k+(-1)^n \frac{x^n}{1+x} \end{aligned} and so integrating over the interval [0,1] gives

\displaystyle \ln 2 = \sum_{k=1}^n \frac{(-1)^{k-1}}{k}+(-1)^n \int_0^1 \frac{x^n}{1+x} \ dx.

Thus \displaystyle \left|\ln 2 + \sum_{k=1}^n \frac{(-1)^k}{k} \right|=\int_0^1 \frac{x^n}{1+x} \ dx and hence

\displaystyle \lim_{n\to\infty} n \left|\ln 2 + \sum_{k=1}^n \frac{(-1)^k}{k} \right|=\lim_{n\to\infty} \int_0^1 \frac{x^n}{1+x} \ dx = \frac{1}{2},

by the second part of the problem in this post. The result now follows from the above problem. \Box

Exercise. Show that the result given in the above problem
i) is true if a=-\infty and the sequence \{1+a_n\} is eventually non-negative.
ii) may or may not be true if \{1+a_n\} is not eventually non-negative.

Integral of cos(ax)e^(-x^2) and integral of sin(ax)e^(-x^2)

Problem 1. i) Show that

i) \displaystyle  \int_0^{\infty}x^{2n}e^{-x^2}dx=\frac{(2n)! \sqrt{\pi}}{2^{2n+1}n!} for all integers n \ge 0.

ii) \displaystyle \int_0^{\infty}\cos(ax) e^{-x^2} dx = \frac{\sqrt{\pi}}{2}e^{-\frac{a^2}{4}} for all real numbers a.

Solution. i) Let \displaystyle I_n:=\int_0^{\infty}x^{2n}e^{-x^2}dx. We showed here that \displaystyle I_0=\int_0^{\infty} e^{-x^2}=\frac{\sqrt{\pi}}{2}. Now, in I_n, we apply integration by parts with \displaystyle x^{2n-1}=u and \displaystyle xe^{-x^2}dx=dv to get \displaystyle I_n=\frac{2n-1}{2}I_{n-1} and thus

\displaystyle \begin{aligned} I_n=\frac{2n-1}{2} \cdot \frac{2n-3}{2}I_{n-2} = \cdots = \frac{2n-1}{2} \cdot \frac{2n-3}{2} \cdots \frac{1}{2} \cdot I_0 \\ =\frac{(2n)!}{2^n(2n) (2n-2) \cdots 2}I_0 =\frac{(2n)! \sqrt{\pi}}{2^{2n+1}n!}. \end{aligned}

ii) Using i) and the Maclaurin series of \cos(ax), we have

\displaystyle \begin{aligned} \int_0^{\infty}\cos(ax) e^{-x^2}\ dx= \int_0^{\infty}\sum_{n=0}^{\infty}(-1)^n\frac{(ax)^{2n}}{(2n)!}e^{-x^2}dx=\sum_{n=0}^{\infty} \frac{(-a^2)^n}{(2n)!} \int_0^{\infty}x^{2n}e^{-x^2}dx \\ =\sum_{n=0}^{\infty} \frac{(-a^2)^n}{(2n)!} \cdot \frac{(2n)! \sqrt{\pi}}{2^{2n+1}n!} =\frac{\sqrt{\pi}}{2}\sum_{n=0}^{\infty}\frac{(\frac{-a^2}{4})^n}{n!} \\ =\frac{\sqrt{\pi}}{2}e^{\frac{-a^2}{4}}. \ \Box \end{aligned}

Unlike \displaystyle \int_0^{\infty}\cos(ax) e^{-x^2} dx, there’s no closed form for \displaystyle \int_0^{\infty}\sin(ax) e^{-x^2} dx but still there’s something nice and non-trivial that we can prove about the integral, as the following problem shows.

Problem 2. i) Show that

i) \displaystyle  \int_0^{\infty}x^{2n+1}e^{-x^2}dx=\frac{n!}{2} for all integers n \ge 0.

ii) \displaystyle \int_0^{\infty}\sin(ax) e^{-x^2} dx = e^{\frac{-a^2}{4}} \int_0^{\frac{a}{2}}e^{x^2}dx for all real numbers a.

Solution. i) Let \displaystyle J_n:=\int_0^{\infty}x^{2n+1}e^{-x^2}dx. See that \displaystyle J_0=\int_0^{\infty}xe^{-x^2}dx=\frac{1}{2}. Integration by parts with x^{2n}=u and xe^{-x^2}dx=dv gives

\displaystyle J_n=nJ_{n-1}=n(n-1)J_{n-2}= \cdots = n(n-1) \cdots 2 \cdot 1 \cdot J_0=\frac{n!}{2}.

ii) Using i) and the Maclaurin series of \sin(ax), we have

\displaystyle \begin{aligned} \int_0^{\infty}\sin(ax) e^{-x^2} dx= \int_0^{\infty}\sum_{n=0}^{\infty}(-1)^n\frac{(ax)^{2n+1}}{(2n+1)!}e^{-x^2}dx \\ =a\sum_{n=0}^{\infty} \frac{(-a^2)^n}{(2n+1)!} \int_0^{\infty}x^{2n+1}e^{-x^2}dx =\frac{a}{2}\sum_{n=0}^{\infty} \frac{(-a^2)^{n}n!}{(2n+1)!} \\ = \frac{a}{2}\sum_{n=0}^{\infty} \frac{(\frac{-a^2}{4})^n4^n}{(2n+1)\binom{2n}{n}n!}. \end{aligned} \ \ \ \ \ \ \ \ \ \ \ \ (*)

But we showed here that \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1}x \ dx = \frac{4^n}{(2n+1)\binom{2n}{n}}. Thus the substitution x=\cos t gives

\displaystyle \int_0^1 (1-x^2)^ndx = \int_0^{\frac{\pi}{2}} \sin^{2n+1}t \ dx=\frac{4^n}{(2n+1)\binom{2n}{n}}.

So, by (*),

\displaystyle \begin{aligned} \int_0^{\infty}\sin(ax) e^{-x^2} dx= \frac{a}{2}\sum_{n=0}^{\infty} \frac{(\frac{-a^2}{4})^n}{n!}\int_0^1(1-x^2)^n dx=\frac{a}{2}\int_0^1 \sum_{n=0}^{\infty} \frac{(\frac{-a^2(1-x^2)}{4})^n}{n!}dx \\ =\frac{a}{2} \int_0^1 e^{\frac{-a^2(1-x^2)}{4}}dx =\frac{a}{2}e^{\frac{-a^2}{4}}\int_0^1 e^{\frac{a^2x^2}{4}}dx \\ = e^{\frac{-a^2}{4}} \int_0^{\frac{a}{2}}e^{x^2}dx. \ \Box \end{aligned}

Exercise 1. In Problem 2, ii), we showed that \displaystyle \frac{1}{2}\sum_{n=0}^{\infty}(-1)^n\frac{a^{2n+1}n!}{(2n+1)!}= e^{\frac{-a^2}{4}} \int_0^{\frac{a}{2}}e^{x^2}dx, for all real numbers a. Here’s another way to prove it. Let

\displaystyle f(a):=\frac{1}{2}e^{\frac{a^2}{4}}\sum_{n=0}^{\infty}(-1)^n\frac{a^{2n+1}n!}{(2n+1)!}- \int_0^{\frac{a}{2}}e^{x^2}dx.

Show that f'(a)=0 and conclude that f(a)=0.

Exercise 2. Evaluate \displaystyle \lim_{a\to\infty} \int_0^{\infty} \sin(ax)e^{-x^2}dx and \displaystyle \lim_{a\to\infty} a \int_0^{\infty} \sin(ax)e^{-x^2}dx.