Integral inequalities (6)

Problem. Let f: [a,b] \longrightarrow \mathbb{R} be a continuously differentiable function and suppose that \displaystyle \int_a^bf(x) \ dx=0. Show that

\displaystyle \int_a^b(f(x))^2dx \le \frac{(b-a)^2}{2}\int_a^b(f'(x))^2 dx.

Solution. Let

\displaystyle g(x):=\int_a^x f(t) \ dt, \ \ \ x \in [a,b].

Then \displaystyle \int_a^b(f(x))^2dx=\int_a^bf(x)g'(x) \ dx and so integration by parts with f(x)=u, \ \ g'(x) \ dx=dv together with the fact that g(a)=g(b)=0, give

\displaystyle \int_a^b(f(x))^2dx=-\int_a^bf'(x)g(x) \ dx.

Thus, by Cauchy-Schwarz,

\displaystyle \left(\int_a^b(f(x))^2dx\right)^2=\left(\int_a^bf'(x)g(x) \ dx\right)^2 \le \int_a^b(f'(x))^2dx \int_a^b(g(x))^2dx. \ \ \ \ \ \ \ \ \ \ (1)

On the other hand, by Cauchy-Schwarz again,

\displaystyle (g(x))^2=\left(\int_a^xf(t) \ dt\right)^2 \le \int_a^x(f(t))^2dt \int_a^xdt \le (x-a)\int_a^b(f(t))^2dt

and hence

\displaystyle \int_a^b(g(x))^2dx \le \int_a^b(x-a) \ dx \int_a^b(f(t))^2dt=\frac{(b-a)^2}{2}\int_a^b(f(t))^2dt. \ \ \ \ \ \ \ \ \ (2)

The result now follows from (1),(2). \ \Box

Exercise 1. Explain why, in the solution of the above problem, the result follows from (1),(2) ?
Hint. If \displaystyle \int_a^b(f(x))^2 dx=0, then f(x)=0 for all x \in [a,b].

Exercise 2. Let f: [a,b] \longrightarrow \mathbb{R} be a continuously differentiable function and suppose that \displaystyle f(a)f(b)=0. Show that

\displaystyle \int_a^b(f(x))^2dx \le \frac{(b-a)^2}{2}\int_a^b(f'(x))^2 dx.

Hint. If f(a)=0, then

\displaystyle (f(x))^2=\left(\int_a^xf'(t) \ dt\right)^2 \le \int_a^x(f'(t))^2dt \int_a^xdt.

Exercise 3. This exercise generalizes the above problem. Let f: [a,b] \longrightarrow \mathbb{R} be a continuously differentiable function. Show that

\displaystyle \int_a^b(f(x))^2dx \le \frac{1}{b-a}\left(\int_a^bf(x) \ dx\right)^2+\frac{(b-a)^2}{2}\int_a^b(f'(x))^2 dx.

Hint. Apply the result given in the problem to the function

\displaystyle F(x):=f(x)-\frac{1}{b-a}\int_a^bf(x) \ dx.

Exercise 4. Prove the inequality given in Exercise 3 for f(x):=\tan^{-1}x, \ \ x \in [0,1], i.e. by evaluating each integral, show that

\displaystyle \int_0^1(\tan^{-1}x)^2 dx \le \left(\int_0^1\tan^{-1}x \ dx\right)^2+\frac{1}{2}\int_0^1\frac{dx}{(x^2+1)^2}.

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If lim (ra_{n+1}-a_n)=L, then lim a_n=L/(r-1)

Problem. Let \{a_n\} be a sequence of real numbers and let r be a real constant.
Show that if |r| > 1 and \displaystyle \lim_{n\to\infty}(ra_{n+1}-a_n)=\ell, for some real number \ell, then \displaystyle \lim_{n\to\infty}a_n=\frac{\ell}{r-1}.

Solution. Let

\displaystyle b_n:=a_n-\frac{\ell}{r-1}.

We are given that \displaystyle \lim_{n\to\infty}(rb_{n+1}-b_n)=0 and we want to show that \displaystyle \lim_{n\to\infty}b_n=0. Using telescoping sum, we can write

\displaystyle b_n=\sum_{k=1}^{n-1}\frac{1}{r^k}(rb_{n-k+1}-b_{n-k})+\frac{b_1}{r^{n-1}}. \ \ \ \ \ \ \ \ \ \ \ (1)

Let \epsilon > 0 be given. Since \displaystyle \lim_{n\to\infty}(rb_{n+1}-b_n)=0, there exist an integer N > 0 and M \ge 0 such that

\begin{aligned}|rb_{n+1}-b_n| < \epsilon, \ \ \ \ \forall n \ge N \ \ \ \ \ \ \ \ \ (2), \\ \\ |rb_{n+1}-b_n| \le M, \ \ \ \ \forall n \ \ \ \ \ \ \ \ \ \ (3).\end{aligned}

Since |r| > 1, we can choose N large enough to also have

\displaystyle \frac{1}{|r|^N} < \epsilon. \ \ \ \ \ \ \ \ \ \ \ (4)

Now, by (1), for n \ge 2N we have

\displaystyle \begin{aligned}|b_n|=\left|\sum_{k=1}^N \frac{1}{r^k}(rb_{n-k+1}-b_{n-k})+\sum_{k=N+1}^{n-1}\frac{1}{r^k}(rb_{n-k+1}-b_{n-k})+\frac{b_1}{r^{n-1}}\right| \\ \le \sum_{k=1}^N \frac{1}{|r|^k}|rb_{n-k+1}-b_{n-k}|+\sum_{k=N+1}^{n-1}\frac{1}{|r|^k}|rb_{n-k+1}-b_{n-k}|+\frac{|b_1|}{|r|^{n-1}} \\ < \epsilon \sum_{k=1}^N \frac{1}{|r|^k}+M\sum_{k=N+1}^{n-1}\frac{1}{|r|^k}+\frac{|b_1|}{|r|^{n-1}}, \ \ \ \ \ \ \ \ \ \ \text{by} \ (2), (3) \\ < \epsilon\sum_{k=1}^{\infty}\frac{1}{|r|^k}+M\sum_{k=N+1}^{\infty}\frac{1}{|r|^k}+\frac{|b_1|}{|r|^N} \\ =\frac{\epsilon}{|r|-1}+\frac{M}{|r|^N(|r|-1)}+\frac{|b_1|}{|r|^N} < \left(\frac{M+1}{|r|-1}+|b_1|\right)\epsilon, \ \ \ \ \ \ \ \text{by} \ (4),\end{aligned}

and so \displaystyle \lim_{n\to\infty}b_n=0. \ \Box

Exercise 1. The result given in the above problem is obviously true for r=0. Show that the result is not necessarily true for 0 \ne |r| \le 1.
Hint. For example, consider the sequence \displaystyle a_n:=\frac{1}{r^n}.

Exercise 2. Let’s take a closer look at the case r=1 in the above problem. Let \displaystyle \lim_{n\to\infty}(a_{n+1}-a_n)=\ell for some real number \ell. What are the possible values of \displaystyle \lim_{n\to\infty}a_n ? Give an example for each possible value of \displaystyle \lim_{n\to\infty}a_n.
Hint. Consider two cases: \ell=0 and \ell \ne 0. You might find Stolz–Cesàro helpful.

Exercise 3. Let \{a_n\} be a sequence of real numbers and let r be a real constant. If \displaystyle \lim_{n\to\infty}(ra_{n+1}-a_n)=\pm \infty, then what can we say about \displaystyle \lim_{n\to\infty}a_n ?

Jensen’s inequality (1)

I will discuss a more general case of Jensen’s inequality in the second part of this post but for now let’s see a special case of the inequality which also gives the main idea of the proof for the general case.

Problem. Let \displaystyle f: [a,b] \longrightarrow (0,\infty) be a continuous function. Show that

\displaystyle \ln\left(\frac{1}{b-a}\int_a^b f(x) \ dx \right) \ge \frac{1}{b-a}\int_a^b \ln(f(x)) \ dx

with equality if and only if f is constant.

Solution. If f is constant, then it’s clear that

\displaystyle \ln\left(\frac{1}{b-a}\int_a^b f(x) \ dx \right) = \frac{1}{b-a}\int_a^b \ln(f(x)) \ dx=\ln(f(x)).

So suppose that f is not constant. Let

\displaystyle c:= \frac{1}{b-a}\int_a^b f(x) \ dx, \ \ \  h(x):=f(x)-c

and see that

\displaystyle \int_a^b h(x) \ dx = 0.

Now, we have

\displaystyle \begin{aligned} \ln(f(x))= \ln(h(x)+c)=\ln c + \ln \left(\frac{1}{c}h(x)+1\right) \le \ln c + \frac{1}{c}h(x). \end{aligned} \ \ \ \ \ \ \ \ \ (*)

Since the only solution of \ln(t+1)=t is t=0, the inequality in (*) becomes equality only for those x for which h(x)=0. Since f is not a constant function, h is not identically zero on [a,b] and so the inequality in (*) is not equality for all x \in [a,b]. Thus integrating (*) gives

\displaystyle \int_a^b \ln(f(x)) \ dx < (b-a)\ln c +\frac{1}{c}\int_a^b h(x) \ dx = (b-a)\ln c

and the result follows. \Box

Example. Since e^{-x^2} is decreasing for x > 0, it is clear that \displaystyle \int_0^t e^{-x^2}dx > te^{-t^2} for all t > 0. Find the smallest real number \alpha such that \displaystyle \int_0^t e^{-x^2}dx > te^{-\alpha t^2} for all t > 0.

Solution. Applying Jensen’s inequality to f(x):=e^{-x^2} on the interval [0,t] we have

\displaystyle \begin{aligned} -\ln t+\ln\left(\int_0^te^{-x^2}dx\right)=\ln\left(\frac{1}{t}\int_0^te^{-x^2} dx\right) > \frac{1}{t}\int_0^t\ln(e^{-x^2}) \ dx = -\frac{t^2}{3}\end{aligned}

and thus

\displaystyle \int_0^te^{-x^2}dx > te^{-t^2/3}. \ \ \ \ \ \ \ \ \ \ \ (\dagger)

Now suppose that \alpha satisfies the inequality \displaystyle \int_0^t e^{-x^2}dx > te^{-\alpha t^2} for all t > 0. Then, using the power series expansion of e^{-x^2}, we have

\displaystyle t-\frac{t^3}{3}+\frac{t^5}{10}- \cdots > t\left(1-\alpha t^2+\frac{\alpha^2t^4}{2} - \cdots \right),

which gives

\displaystyle \frac{1}{3}-\frac{t^2}{10} + \cdots < \alpha - \frac{\alpha^2t^2}{2} + \cdots

and so, letting t \to 0^+, we get \displaystyle \alpha \ge \frac{1}{3}. Hence, by \displaystyle (\dagger), \ \alpha=\frac{1}{3} is the smallest real number for which the inequality \displaystyle \int_0^t e^{-x^2}dx > te^{-\alpha t^2} holds for all t > 0. \ \Box

Integral of ln(x)/((x-1)(x+a))

Problem. Given a real constant a > 0, show that

i) \displaystyle \int_0^1 \left(\frac{1}{x+a}+\frac{a}{1+ax}\right)\ln x \ dx = -\frac{\pi^2}{6}-\frac{1}{2}(\ln a)^2

ii) \displaystyle \int_0^{\infty}\frac{\ln x}{(x-1)(x+a)} \ dx=\frac{\pi^2+(\ln a)^2}{2(a+1)}.

Solution. i) Let

\displaystyle f(a):=\int_0^1 \left(\frac{1}{x+a}+\frac{a}{1+ax}\right)\ln x \ dx

and write

\displaystyle f(a)=\int_0^1 \frac{\ln x}{x+a} \ dx + \int_0^1 \frac{a \ln x}{1+ax} \ dx.

Now in the first integral on the right-hand side of the above , change x to ax and in the second integral, change ax to x to get

\displaystyle f(a)=-(\ln a)^2+\int_0^{1/a} \frac{\ln x}{x+1} \ dx + \int_0^a \frac{\ln x}{x+1} \ dx.

Thus differentiating with respect to a gives \displaystyle f'(a)=-\frac{\ln a}{a} and hence integrating gives

\displaystyle f(a)=-\frac{1}{2}(\ln a)^2 + C, \ \ \ \ \ \ \ \ \ \ \ (*)

for some constant C. But since, by the definition of f,

\displaystyle f(1)=2\int_0^1 \frac{\ln x}{x+1} \ dx = -\frac{\pi^2}{6},

we have, by \displaystyle (*), \ C=f(1)=-\frac{\pi^2}{6} and hence \displaystyle f(a)=-\frac{\pi^2}{6}-\frac{1}{2}(\ln a)^2.

ii) Write

\displaystyle \int_0^{\infty}\frac{\ln x}{(x-1)(x+a)} \ dx=\int_0^1 \frac{\ln x}{(x-1)(x+a)} \ dx+\int_1^{\infty}\frac{\ln x}{(x-1)(x+a)} \ dx

and so after changing x to \displaystyle \frac{1}{x} in the second integral on the right-hand side of the above, we get

\displaystyle \begin{aligned}\int_0^{\infty}\frac{\ln x}{(x-1)(x+a)} \ dx= \int_0^1 \frac{\ln x}{(x-1)(x+a)} \ dx+\int_0^1 \frac{\ln x}{(x-1)(1+ax)} \ dx \\ =\frac{2}{a+1}\int_0^1 \frac{\ln x}{x-1} \ dx- \frac{1}{a+1}\int_0^1\left(\frac{1}{x+a}+\frac{a}{1+ax}\right) \ln x \ dx \\ =\frac{\pi^2}{3(a+1)}-\frac{1}{a+1}\int_0^1\left(\frac{1}{x+a}+\frac{a}{1+ax}\right) \ln x \ dx=\frac{\pi^2+(\ln a)^2}{2(a+1)},\end{aligned}

by i). \Box

Exercise 1. In the solution of the above problem, we used the identities

\displaystyle \int_0^1 \frac{\ln x}{x-1} \ dx = \frac{\pi^2}{6}, \ \ \ \ \int_0^1 \frac{\ln x}{x+1} \ dx =-\frac{\pi^2}{12}

without proving them. Prove them!
Hint. Make the substitution \ln x = -t and then see Problem 2 in this post!

Exercise 2. In the solution of the first part of the above problem, I claimed, without proof, that \displaystyle f'(a)=-\frac{\ln a}{a}. Prove the claim!

Exercise 3. Using convergence tests, show that \displaystyle \int_0^{\infty}\frac{\ln x}{(x-1)(x+a)} \ dx is convergent for all real constants a > 0.

Bounds for the integral of (sin(x))^(sin(x)) on the interval [0,pi/2]

Problem. Show that

\displaystyle 1.26 < \int_0^{\pi/2}(\sin x)^{\sin x} dx < 1.31.

Solution. Since (\sin x)^{\sin x}=e^{\sin x \ln(\sin x)} and

\displaystyle 1+t \le e^t \le 1+t+\frac{t^2}{2},

for all t \le 0, by Taylor’s theorem, we have

\displaystyle 1+\sin x \ln(\sin x) \le (\sin x)^{\sin x} \le 1+ \sin x \ln(\sin x) +\frac{1}{2}(\sin x \ln(\sin x))^2, \ \ \ \ \ \ \ \ \ \ (1)

for all x \in (0,\pi/2]. Now let

\displaystyle \begin{aligned} I:=\int_0^{\pi/2}(\sin x)^{\sin x} \ dx, \ \ J:=\int_0^{\pi/2}\sin x \ln(\sin x) \ dx, \ \ \ K:=\int_0^{\pi/2}(\sin x \ln(\sin x))^2 dx.\end{aligned}

So by (1),

\displaystyle \frac{\pi}{2}+J \le I \le \frac{\pi}{2} + J+\frac{1}{2}K. \ \ \ \ \ \ \ \ \ (2)

So we need to find J and K. Let’s find J first.

We (carefully) use integration by parts with \sin x \ dx = dv and \ln(\sin x) = u. Then v=1-\cos x and \displaystyle du=\frac{\cos x}{\sin x} \ dx and thus

\displaystyle \begin{aligned}J=-\int_0^{\pi/2}\frac{\cos x(1-\cos x)}{\sin x} \ dx=-\int_0^{\pi/2}\frac{\cos x \sin x}{1+\cos x} \ dx=-\int_0^1\frac{x}{1+x} \ dx=\ln 2 -1.\end{aligned} \ \ \ \ \ \ (3)

So, by (2),

\displaystyle I \ge \frac{\pi}{2}+J=\frac{\pi}{2}+\ln 2 - 1 \approx 1.2639 > 1.26.

Now we find K. Using integration by parts with \sin^2x \ dx = dv, \ (\ln(\sin x))^2 = u, we have \displaystyle v=\frac{1}{2}(x-\sin x \cos x) and du=2\cot x \ln(\sin x) \ dx and hence

\displaystyle \begin{aligned}K=\int_0^{\pi/2}(-x+\sin x \cos x) \cot x \ln(\sin x) \ dx.\end{aligned}

So if we put

\displaystyle K_1:=\int_0^{\pi/2}\cos^2x \ln(\sin x) \ dx, \ \ \ \ K_2:= \int_0^{\pi/2} x \cot x \ln(\sin x) \ dx,

then

K=K_1-K_2. \ \ \ \ \ \ \ \ \ \ (4)

To find K_1, we use integration by parts again with \cos^2x \ dx = dv, \ u=\ln(\sin x) to get

\displaystyle K_1=-\frac{1}{2}\int_0^{\pi/2}(x \cot x + \cos^2 x) \ dx=-\frac{\pi}{8}-\frac{1}{2}\int_0^{\pi/2}x \cot x \ dx.

Again, integration by parts with \cot x \ dx = dv, \ x=u together with the fact that \displaystyle \int_0^{\pi/2}\ln(\sin x) \ dx = -\frac{\pi}{2}\ln 2 (see this post!) give \displaystyle \int_0^{\pi/2}x \cot x \ dx = \frac{\pi}{2}\ln 2 and so

\displaystyle K_1=-\frac{\pi}{8}-\frac{\pi}{4}\ln 2. \ \ \ \ \ \ \ \ \ \ (5)

And finally, to evaluate K_2, we use integration by parts again, this time with \cot x \ln(\sin x) \ dx = dv and x =u to get

\displaystyle K_2=-\frac{1}{2}\int_0^{\pi/2}(\ln(\sin x))^2 \ dx = -\frac{\pi^3}{48}-\frac{\pi}{4}(\ln 2)^2. \ \ \ \ \ \ \ \ \ \ (6)

For two proofs of the identity \displaystyle \int_0^{\pi/2}(\ln(\sin x))^2 \ dx = \frac{\pi^3}{24}+\frac{\pi}{2}(\ln 2)^2, see this post!

So by (4),(5) and (6),

\displaystyle K=K_1-K_2=-\frac{\pi}{8}-\frac{\pi}{4}\ln 2+\frac{\pi^3}{48}+\frac{\pi}{4}(\ln 2)^2 \ \ \ \ \ \ \ \ \ (7)

and thus, by (2), (3), (7)

\displaystyle \begin{aligned} I \le \frac{\pi}{2} + J+\frac{1}{2}K=\frac{7\pi}{16}-1+\ln 2 -\frac{\pi}{8}\ln 2+\frac{\pi}{8}(\ln 2)^2+\frac{\pi^3}{96} \approx 1.307 < 1.31. \ \Box\end{aligned}

Remark. According to Wolfram, \displaystyle \int_0^{\pi/2}(\sin x)^{\sin x} dx \approx 1.302944.

Exercise 1. Try to fully understand the integration by parts we used in the solution of the above problem to evaluate J.

Exercise 2. Show that \displaystyle 2.52 < \int_0^{\pi}(\sin x)^{\sin x} dx < 2.62.

The alternating series \sum (-1)^n*ln(n)/n and the integral of ln(x)/(1+e^x)

Problem. Let \gamma be the Euler’s constant. Show that

i) \displaystyle 0 < n\left(\ln 2 - \sum_{k=1}^n \frac{1}{n+k}\right) < 1 for all integers n \ge 1

ii) \displaystyle \sum_{n=1}^{\infty} (-1)^n \ \frac{\ln n}{n}=\gamma \ln 2 - \frac{1}{2}(\ln 2)^2

iii) \displaystyle \int_0^{\infty} \frac{\ln x}{1+e^x} \ dx = -\frac{1}{2}(\ln 2)^2.

Solution. i) Consider the sequences \displaystyle a_n:=\sum_{k=1}^n \frac{1}{n+k} and \displaystyle b_n:=a_n+\frac{1}{n}. Then

\displaystyle a_{n+1}-a_n=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}=\frac{1}{(2n+1)(2n+2)} > 0

and

\displaystyle b_{n+1}-b_n=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n}=-\frac{3n+2}{n(2n+1)(2n+2)} < 0.

So \{a_n\} is increasing, \{b_n\} is decreasing and

\displaystyle \lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}=\int_0^1 \frac{dx}{1+x}=\ln 2.

Thus a_n < \ln 2 < b_n for all n \ge 1 and the result follows.

ii) We first need to show that the series converges. That is easy to see; differentiating shows that the function \displaystyle y=\frac{\ln x}{x} is decreasing for x \ge e and clearly \displaystyle \lim_{x\to\infty}y=0, So the series converges by the alternating test.
Now let

\displaystyle S:=\sum_{n=1}^{\infty} (-1)^n \ \frac{\ln n}{n}, \ \ \ S_n= \sum_{k=1}^n (-1)^k \ \frac{\ln k}{k}.

Since S converges, we have

\displaystyle S=\lim_{n\to\infty} S_{2n}. \ \ \ \ \ \ \ \ \ \ (1)

We also have

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

and thus, by (1),

\displaystyle \begin{aligned} S=\ln 2 \lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k} -\ln n\right) + \lim_{n\to\infty}\ln n \left(\ln 2 - \sum_{k=1}^n \frac{1}{n+k}\right)-\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \frac{\ln(1+\frac{k}{n})}{1+\frac{k}{n}} \\ =\gamma \ln 2 + \lim_{n\to\infty}\ln n \left(\ln 2 - \sum_{k=1}^n \frac{1}{n+k}\right) - \int_0^1 \frac{\ln(1+x)}{1+x} \ dx \\ = \gamma \ln 2 + \lim_{n\to\infty}\ln n \left(\ln 2 - \sum_{k=1}^n \frac{1}{n+k}\right)-\frac{1}{2}(\ln 2)^2. \ \ \ \ \ \ \ (2) \end{aligned}

Finally, by the first part of the problem,

\displaystyle 0 <  \ln n \left(\ln 2 - \sum_{k=1}^n \frac{1}{n+k}\right) < \frac{\ln n}{n}

and hence, by the squeeze theorem, \displaystyle \lim_{n\to\infty}\ln n \left(\ln 2 - \sum_{k=1}^n \frac{1}{n+k}\right)=0. Therefore \displaystyle S=\gamma \ln 2 - \frac{1}{2}(\ln 2)^2, by (2).

iii) We have

\displaystyle \begin{aligned} \int_0^{\infty} \frac{\ln x}{1+e^x} \ dx = \int_0^{\infty} \frac{e^{-x}\ln x}{1+e^{-x}} \ dx=\int_0^{\infty}e^{-x}\ln x \sum_{n=0}^{\infty}(-1)^ne^{-nx} \ dx \\ =\sum_{n=0}^{\infty}(-1)^n \int_0^{\infty}e^{-(n+1)x}\ln x \ dx=\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}\int_0^{\infty}e^{-x}\ln \left(\frac{x}{n+1}\right)dx \\ =\sum_{n=1}^{\infty}\frac{(-1)^n}{n+1}\left(\int_0^{\infty}e^{-x}\ln x \ dx - \ln(n+1) \int_0^{\infty}e^{-x} dx\right) \\ =\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}(-\gamma - \ln(n+1))=-\gamma \sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}+\sum_{n=0}^{\infty}(-1)^{n+1} \ \frac{\ln(n+1)}{n+1} \\ =-\gamma \ln 2 +\sum_{n=1}^{\infty}(-1)^n \ \frac{\ln n}{n}=-\frac{1}{2}(\ln 2)^2, \ \ \ \ \text{by ii)}. \ \Box \end{aligned}

Remark. At the end of the solution of the second part of the above problem, we used the first part of the problem to prove that \displaystyle \lim_{n\to\infty}\ln n \left(\ln 2 - \sum_{k=1}^n \frac{1}{n+k}\right)=0. That can also be done using the result we proved in this post. Here’s how. Let \displaystyle f(x):=\frac{1}{1+x} and see that

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

and so

\displaystyle \begin{aligned} \lim_{n\to\infty} \ln n \left(\ln 2 - \sum_{k=1}^n \frac{1}{n+k}\right)=\lim_{n\to\infty} \frac{\ln n}{n} \cdot \frac{f(0)-f(1)}{2}=\lim_{n\to\infty} \frac{\ln n}{n} \cdot \frac{1}{4}=0.\end{aligned}

Using IVT and MVT (4)

Problem. Let \displaystyle f:[a,b] \longrightarrow \mathbb{R} be a continuous function which is differentiable on (a,b). Suppose also that f(a)=a, \ f(b)=b. Show that for any positive real numbers r_1, \cdots , r_n, there exist distinct elements x_1, \cdots , x_n \in (a,b) such that

\displaystyle \sum_{i=1}^n \frac{r_i}{f'(x_i)}=\sum_{i=1}^n r_i.

Solution. By the intermediate value theorem, for each 1 \le i \le n-1, there exists c_i \in (a,b) such that

\displaystyle f(c_i)=a+\frac{r_1+ \cdots r_i}{r_1+\cdots + r_n}(b-a), \ \ \ \ \ \ \ \ \ (1)

because f(a)=a, \ f(b)=b, and \displaystyle a+\frac{r_1+ \cdots r_i}{r_1+\cdots + r_n}(b-a) \in (a,b), for all 1 \le i \le n-1.
Let’s define c_0:=a, \ c_n:=b. Notice that now (1) holds for 0 \le i \le n. Also notice that, by relabeling c_i if necessary, we may assume that c_1 < c_2 < \cdots < c_{n-1}.
Now, by the mean value theorem, for each 1 \le i \le n, there exists x_i \in (c_{i-1},c_i) such that

f(c_i)-f(c_{i-1})=(c_i-c_{i-1})f'(x_i). \ \ \ \ \ \ \ \ \ \ (2)

We also have from (1) that

\displaystyle f(c_i)-f(c_{i-1})=\frac{r_i(b-a)}{r_1+ \cdots + r_n} \ \ \ \ \ \ \ \ \ \ (3)

for all 1 \le i \le n. So, by (2), (3),

\displaystyle \begin{aligned}\sum_{i=1}^n \frac{r_i}{f'(x_i)}=\sum_{i=1}^n \frac{r_i(c_i-c_{i-1})}{f(c_i)-f(c_{i-1})}=\frac{r_1+ \cdots + r_n}{b-a}\sum_{i=1}^n(c_i-c_{i-1}) =\frac{r_1+ \cdots + r_n}{b-a}(c_n-c_0) \\ =\frac{r_1+ \cdots + r_n}{b-a}(b-a)=r_1+\cdots + r_n. \ \Box\end{aligned}

Remark. The result given in the above problem does not necessarily hold if r_i are not all positive. For example, consider the function f(x)=x^2, \ x \in [0,1], with n=2 and r_1=1, \ r_2=-1. There are no x_1 \neq x_2 such that \displaystyle \frac{1}{2x_1}-\frac{1}{2x_2}=0.

Example. Let f: [a,b] \longrightarrow \mathbb{R} be a continuous function such that \displaystyle \int_a^b f(x) \ dx \ne 0. Show that for any positive real numbers r_1, \cdots , r_n, there exist distinct elements x_1, \cdots , x_n \in (a,b) such that

\displaystyle \sum_{i=1}^n \frac{r_i}{f(x_i)}=\frac{b-a}{\int_a^bf(x) \  dx}\sum_{i=1}^nr_i.

Solution. Let

\displaystyle g(x):=a+\frac{b-a}{\int_a^bf(x) \ dx}\int_a^x f(t) \ dt, \ \ \ \ x \in [a,b].

Then g(a)=a, \ g(b)=b and \displaystyle g'(x)=\frac{b-a}{\int_a^bf(x) \ dx}f(x). Now apply the result given in the above problem to g. \ \Box

Limit of integrals (18)

The following problem looks weird at first but it’s really not. The problem basically comes from my (successful) attempt to prove the following nice-looking limit in multi-variable calculus

\displaystyle \lim_{n\to\infty} \int_{0}^{1} \cdots \int_{0}^{1} \frac{x_1 + \cdots + x_n}{x_1^2 + \cdots + x_n^2} \ dx_1 \ \cdots \ dx_n=\frac{3}{2}.

I have explained that in the remark below the problem.

Problem. Show that \displaystyle \lim_{n\to\infty} n\int_0^{\infty}\frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^xe^{-t^2} dt \right)^n dx=\frac{3}{2}.

Solution. First notice that, since \displaystyle \int_0^{\infty}e^{-t^2}dt=\frac{\sqrt{\pi}}{2} (see this post!), we have

\displaystyle \begin{aligned} 0 < n\int_1^{\infty}\frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^x e^{-t^2} dt  \right)^n dx < n \int_1^{\infty} \frac{1}{x^{n+1}}\left(\int_0^{\infty}e^{-t^2} dt \right)^n dt=\left(\frac{\sqrt{\pi}}{2}\right)^n.\end{aligned}

Hence, since \displaystyle \frac{\sqrt{\pi}}{2} < 1, we get

\displaystyle \lim_{n\to\infty} n\int_1^{\infty}\frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^x e^{-t^2} dt \right)^n dx=0,

by the squeeze theorem, and thus

\displaystyle \lim_{n\to\infty}n\int_0^{\infty}\frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^xe^{-t^2} dt\right)^n dx = \lim_{n\to\infty}n \int_0^1 \frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^x e^{-t^2} dt \right)^n dx. \ \ \ \ \ \ \ \ \ \ \ (1)

Let

\displaystyle I_n:=\int_0^1 \frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^xe^{-t^2} dt\right)^n dx.

Using the Maclaurin series of e^{-x^2}, it’s clear that

\displaystyle x-\frac{x^3}{3} \le \int_0^x e^{-t^2} \ dt \le x-\frac{x^3}{3}+\frac{x^5}{10}

and so

\displaystyle n \int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}\right)^n dx \le nI_n \le n\int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}+\frac{x^4}{10}\right)^n dx. \ \ \ \ \ \ \ \ (2)

We now find the limits of the sequences on the LHS and the RHS of (2). First, the LHS one. The substitution \displaystyle 1-\frac{x^2}{3}=s gives

\displaystyle n \int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}\right)^n dx=\frac{n}{2}\int_{2/3}^1 \frac{1-e^{-3(1-s)}}{1-s}s^n ds

and hence

\displaystyle \lim_{n\to\infty}n \int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}\right)^n dx=\frac{1}{2}\lim_{s\to1}\frac{1-e^{-3(1-s)}}{1-s}=\frac{3}{2} \ \ \ \ \ \ \ \ \ \ \ (3)

because, as we showed here, \displaystyle \lim_{n\to\infty} n\int_c^1 f(x) \ x^n dx =f(1) for any c \in [0,1) and any continuous function f: [0,1] \to \mathbb{R}.

Now we find the limit of the sequence on the RHS of (2). The idea is the same. We first make the substitution x^2=s to get

\displaystyle n\int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}+\frac{x^4}{10}\right)^n dx=\frac{n}{2}\int_0^1 \frac{1-e^{-s}}{s}\left(1-\frac{s}{3}+\frac{s^2}{10}\right)^nds. \ \ \ \ \ \ \ \ \ (4)

The function \displaystyle y=1-\frac{s}{3}+\frac{s^2}{10}, \ \ s \in [0,1], is decreasing and so it has an inverse, i.e. we can find s in terms of y but to avoid the unnecessary mess, let’s just write s=g(y). Then the substitution \displaystyle y=1-\frac{s}{3}+\frac{s^2}{10} and (4) give

\displaystyle \begin{aligned} n\int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}+\frac{x^4}{10}\right)^n dx=-\frac{n}{2}\int_{23/30}^1 \frac{1-e^{-g(y)}}{g(y)}g'(y)y^n dy \\ =-\frac{n}{2}\int_{23/30}^1 \frac{1-e^{-g(y)}}{g(y)}\left(\frac{g(y)}{5}-\frac{1}{3}\right)^{-1}y^ndy\end{aligned}

and so, again using the fact that \displaystyle \lim_{n\to\infty} n\int_c^1 f(x) \ x^n dx =f(1) for any c \in [0,1) and any continuous function f: [0,1] \to \mathbb{R}, we have

\displaystyle \begin{aligned} \lim_{n\to\infty} n\int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}+\frac{x^4}{10}\right)^n dx=-\frac{1}{2}\lim_{y\to1}\frac{1-e^{-g(y)}}{g(y)}\left(\frac{g(y)}{5}-\frac{1}{3}\right)^{-1} \\ =-\frac{1}{2}\lim_{z\to0} \frac{1-e^{-z}}{z}\left(\frac{z}{5}-\frac{1}{3}\right)^{-1}=\frac{3}{2}. \end{aligned} \ \ \ \ \ \ \ \ \ (5)

So (2), (3), (5) and the squeeze theorem together give \displaystyle \lim_{n\to\infty} nI_n=\frac{3}{2} and the result follows from (1). \ \Box

Remark (for those readers who are familiar with multi-variable calculus). It follows from the above problem that

\displaystyle \lim_{n\to\infty} \int_{0}^{1} \cdots \int_{0}^{1} \frac{x_1 + \cdots + x_n}{x_1^2 + \cdots + x_n^2} \ dx_1 \ \cdots \ dx_n=\frac{3}{2}.

That’s because we can write \displaystyle \frac{1}{x_1^2 + \cdots + x_n^2}=\int_0^{\infty} e^{-(x_1^2+ \cdots + x_n^2)x} dx and thus

\displaystyle \begin{aligned} \int_{0}^{1} \cdots \int_{0}^{1} \frac{x_1 + \cdots + x_n}{x_1^2 + \cdots + x_n^2} \ dx_1 \ \cdots \ dx_n= n\int_{0}^{1} \cdots \int_{0}^{1} \frac{x_1}{x_1^2 + \cdots + x_n^2} \ dx_1 \cdots \ dx_n \\ =n\int_{0}^{\infty} \int_0^1 \cdots \int_{0}^{1} x_1e^{-(x_1^2+ \cdots + x_n^2)x} dx_1 \cdots \ dx_n \ dx=n \int_0^{\infty}\left(\int_0^1x_1e^{-x_1^2x}dx_1\right)\left(\int_0^1e^{-t^2x}dt\right)^{n-1}dx \\ =n\int_0^{\infty}\frac{1-e^{-x}}{2x}\left(\int_0^1e^{-t^2x}dt\right)^{n-1}dx=n\int_0^{\infty}\frac{1-e^{-x}}{2x}\left(\int_0^{\sqrt{x}} \frac{e^{-t^2}}{\sqrt{x}} \ dx\right)^{n-1}dx \\ =n\int_0^{\infty}\frac{1-e^{-x^2}}{x^n}\left(\int_0^x e^{-t^2} dt \right)^{n-1} dx \end{aligned}

and the result follows because

\displaystyle \begin{aligned} \lim_{n\to\infty} n\int_0^{\infty}\frac{1-e^{-x^2}}{x^n}\left(\int_0^x e^{-t^2} dt \right)^{n-1} dx=\lim_{n\to\infty} n\int_0^{\infty}\frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^xe^{-t^2} dt \right)^n dx=\frac{3}{2},\end{aligned}

by the above problem.