Improper Integrals: Original Convergence and Evaluation Exercises
Exercise editors and source: Prof. Antonino De Martino and Dr. Luana Manfredini, Eserciziario 2.1. The original exercises and source theory are presented here in English for Logic & Motion.
Improper Integrals: All Three Source Cases
An endpoint singularity or an infinite interval requires a limit of proper integrals. At an interior singularity, both one-sided improper integrals must converge separately; cancellation in a symmetric limit does not establish convergence.
For α > 0:
\[\int_0^\alpha\frac{dx}{x^p}\quad \begin{cases}\text{converges},&p<1,\\\text{diverges},&p\ge1.\end{cases}\] \[\int_\alpha^\infty\frac{dx}{x^p}\quad \begin{cases}\text{converges},&p>1,\\\text{diverges},&p\le1.\end{cases}\]For α > 1:
\[\int_1^\alpha\frac{dx}{(\log x)^p}\quad \begin{cases}\text{converges},&p<1,\\\text{diverges},&p\ge1.\end{cases}\]Absolute comparison and comparison by a finite positive ratio reduce the following original exercises to these cases. Strict inequalities at p = 1 matter.
Original Convergence and Evaluation Problems
Exercise 1
Prove convergence and evaluate the integral:
\[\int_{\frac{1}{e}}^e \frac{\log (|\log x|)}{x} \, dx.\]Solution.
The integrand has an interior logarithmic singularity at x = 1. Split there: \(I=\int_{1/e}^1\frac{\log(-\log x)}x\,dx +\int_1^e\frac{\log(\log x)}x\,dx.\) In the first part use t = −log x and dt = −dx/x; in the second use t = log x and dt = dx/x. Both become the same convergent improper integral: \(\int_{1/e}^1\frac{\log(-\log x)}x\,dx =-\int_1^0\log t\,dt=\int_0^1\log t\,dt.\) \(\int_1^e\frac{\log(\log x)}x\,dx=\int_0^1\log t\,dt.\) \(I=2\lim_{a\to0^+}\left[t\log t-t\right]_a^1 =2\lim_{a\to0^+}(-1-a\log a+a).\)
Final result
\[-2\]Exercise 2
Determine whether the following function is improperly integrable on [1,+∞):
\[f(x)=\frac{\sin x}{x\sqrt{x+1}}.\]Solution.
Use absolute comparison, as in the source: \(0\le\left|\frac{\sin x}{x\sqrt{x+1}}\right| \le\frac1{x\sqrt{x+1}}\le\frac1{x^{3/2}},\qquad x\ge1.\) The last inequality uses √(x+1) ≥ √x. The comparison integral converges because 3/2 > 1. Therefore the original function is integrable in the improper sense, in fact absolutely.
Final result
\[\int_1^\infty\left|\frac{\sin x}{x\sqrt{x+1}}\right|dx<\infty.\]Exercise 3
Determine whether the following function is improperly integrable on [0,1]:
\[f(x)=\frac1{\sqrt{1-x^3}}.\]Solution.
Factor the radicand: \(1-x^3=(1-x)(1+x+x^2).\) For 0 ≤ x < 1: \(0\le\frac1{\sqrt{1-x^3}} =\frac1{\sqrt{1-x}\sqrt{1+x+x^2}} \le\frac1{\sqrt{1-x}}.\) The inequality uses √(1+x+x²) ≥ 1. The comparison integral at the endpoint converges since 1/2 < 1.
Final result
\[\int_0^1\frac{dx}{\sqrt{1-x^3}}<\infty.\]Exercise 4
Prove convergence and evaluate the integral:
\[\int_{2}^{+\infty} \frac{1}{x \sqrt{x^{2}-1}} \,dx.\]Solution.
First check convergence by comparing with 1/x²: \(\lim_{x\to\infty}\frac{1/(x\sqrt{x^2-1})}{1/x^2} =\lim_{x\to\infty}\sqrt{\frac{x^2}{x^2-1}}=1.\) Keep the source’s Euler substitution: \(\sqrt{x^2-1}=x+t,\qquad x=-\frac{t^2+1}{2t}.\) \(dx=-\frac{t^2-1}{2t^2}\,dt, \qquad\sqrt{x^2-1}=\frac{t^2-1}{2t}.\) Both denominator factors are negative multiples of the displayed expressions, so their product cancels the minus sign in dx: \(\int\frac{dx}{x\sqrt{x^2-1}}=2\int\frac{dt}{1+t^2}=2\arctan t+C.\) The lower endpoint is t = √3 − 2. At the upper endpoint: \(t=\sqrt{b^2-1}-b=-\frac1{\sqrt{b^2-1}+b}\longrightarrow0^-.\) Thus: \(I=2\lim_{b\to\infty}\left[\arctan t\right]_{\sqrt3-2}^{\sqrt{b^2-1}-b} =-2\arctan(\sqrt3-2).\) Since √3 − 2 = −tan(π/12), the result is π/6.
Final result
\[\frac\pi6\]Exercise 5
Prove convergence:
\[\int_{0}^{1} \frac{\log(1+\sqrt[4]{x})}{e^{x}-1} \, dx\]Solution.
Compare with x⁻³ᐟ⁴ at zero. The ratio factors into two fundamental limits: \(\lim_{x\to0^+}\frac{\log(1+x^{1/4})/(e^x-1)}{x^{-3/4}} =\lim_{x\to0^+}\frac{x}{e^x-1}\frac{\log(1+x^{1/4})}{x^{1/4}}=1.\) The integrand is continuous away from zero. The comparison integral at zero converges because 3/4 < 1.
Final result
\[\int_0^1\frac{\log(1+\sqrt[4]x)}{e^x-1}\,dx<\infty.\]Exercise 6
Prove convergence and evaluate the integral:
\[\int_{3}^{+ \infty} \frac{x+1}{(x-1)^{2}(x-2)} \, dx.\]Solution.
The integrand is continuous for x ≥ 3. Compute the primitive by Hermite decomposition: \(\frac{x+1}{(x-1)^2(x-2)} =-\frac3{x-1}+\frac3{x-2}-\frac2{(x-1)^2}.\) Equivalently, write the last term as the derivative of 2/(x−1). Matching coefficients gives A = −3, B = 3 and C = 2 in the source’s notation. Hence: \(F(x)=3\log\frac{|x-2|}{|x-1|}+\frac2{x-1}.\) Keep the infinite upper endpoint: \(I=\lim_{b\to\infty}[F(x)]_3^b =\lim_{b\to\infty}\left(3\log\frac{b-2}{b-1}+\frac2{b-1}+3\log2-1\right).\) The first two terms tend to zero.
Final result
\[\log8-1\]Exercise 7
Solve the following exercise:
\[\int_{1}^{+\infty} \frac{1}{(x+1)(x+2)x^{2}} \, dx.\]Solution.
The integrand is continuous and positive on [1,∞) and behaves as x⁻⁴ at infinity, so comparison establishes convergence. Its partial fractions are \(\frac1{x^2(x+1)(x+2)}=-\frac3{4x}+\frac1{2x^2}+\frac1{x+1}-\frac1{4(x+2)}.\) An antiderivative is \(F(x)=-\frac34\log x-\frac1{2x}+\log(x+1)-\frac14\log(x+2).\) The logarithmic leading terms cancel at infinity, so F(∞)=0. At 1 it is −1/2+log 2−(log 3)/4. Subtract this endpoint value.
Final result
\[\frac12-\log2+\frac14\log3\]Exercise 8
Solve the following exercise:
\[\int_{-1}^{1} \frac{1}{(x-4) \sqrt{|x|}} \, dx.\]Solution.
The only interior singularity is zero and behaves as a constant times |x|⁻¹ᐟ², so both one-sided integrals converge. Put t=√x on (0,1] and t=√(−x) on [−1,0). Each half becomes \(I_+=2\int_0^1\frac{dt}{t^2-4},\qquad I_-=-2\int_0^1\frac{dt}{t^2+4}.\) Integrate with partial fractions and the arctangent: \(I_+=\frac12\left[\log\left|\frac{t-2}{t+2}\right|\right]_0^1=-\frac12\log3,\qquad I_-=-[\arctan(t/2)]_0^1.\) Add the convergent contributions.
Final result
\[-\frac12\log3-\arctan\frac12\]