1.6 Practice Problems
Problem 1.6.1. Determine whether each of the following converges, and evaluate it where it does:
- (a)
- \(\displaystyle \int ^{\infty }_{1}\frac {dx}{x^{p}}\), for \(p>0\);
- (b)
- \(\displaystyle \int ^{1}_{0}\frac {dx}{x^{p}}\), for \(p>0\);
- (c)
- \(\displaystyle \int ^{\infty }_{0}\frac {dx}{x^{p}}\), for \(p>0\).
Where to start: (a) and (b) split at \(p=1\) in opposite senses; (c) then follows without any further integration.
Show solution
Solution. (a) converges for \(p>1\) and diverges for \(p\leq 1\), to \(1/(p-1)\) when it converges. (b) converges for \(p<1\) and diverges for \(p\geq 1\), to \(1/(1-p)\).
(c) never converges. Splitting at \(1\) gives the two integrals above, and no single \(p\) makes both converge: (a) needs \(p>1\) and (b) needs \(p<1\). The integrand decays too slowly at infinity when it is well behaved at the origin, and too sharply at the origin when it decays fast enough at infinity. This is the tension that the gamma function of Chapter 2 resolves by using \(e^{-t}\) to control the tail and \(t^{x-1}\) to control the origin separately.
Problem 1.6.2. Show that \(\displaystyle \int ^{\infty }_{0}e^{-ax}\,dx\) converges for \(a>0\) and diverges for \(a\leq 0\), and evaluate it in the convergent case.
Problem 1.6.3. Use the comparison test to determine the convergence of
- (a)
- \(\displaystyle \int ^{\infty }_{1}\frac {dx}{\sqrt {x^{3}+1}}\);
- (b)
- \(\displaystyle \int ^{\infty }_{1}\frac {x\,dx}{x^{3}+2}\);
- (c)
- \(\displaystyle \int ^{\infty }_{2}\frac {dx}{\ln x}\).
Where to start: compare each with a suitable \(x^{-p}\); for (c) note that \(\ln x < x\) for large \(x\).
Problem 1.6.4. Determine whether \(\displaystyle \int ^{1}_{0}\frac {dx}{\sqrt {1-x^{2}}}\) converges, and evaluate it if so. Identify the kind of improper integral it is and where the difficulty lies.
Problem 1.6.5. Show that \(\displaystyle \int ^{\infty }_{0}\frac {\sin x}{x}\,dx\) converges but \(\displaystyle \int ^{\infty }_{0}\left |\frac {\sin x}{x}\right |dx\) does not: the first is convergent but not absolutely convergent. Where to start: for the second, bound the integral over \([k\pi ,(k+1)\pi ]\) below by a multiple of \(1/(k+1)\) and compare with the harmonic series.
Show solution
Solution. Convergence of the first follows from the Dirichlet test: \(1/x\) decreases monotonically to zero and the partial integrals of \(\sin x\) are bounded.
For the second, on \([k\pi ,(k+1)\pi ]\) we have \(x\leq (k+1)\pi \), so \[\int ^{(k+1)\pi }_{k\pi }\left |\frac {\sin x}{x}\right |dx \geq \frac {1}{(k+1)\pi }\int ^{(k+1)\pi }_{k\pi }\left |\sin x\right |dx = \frac {2}{(k+1)\pi }.\] Summing over \(k\) gives a multiple of the harmonic series, which diverges.
The distinction matters. An integral that converges only conditionally cannot be rearranged or split freely, and the interchanges of limit and integral used throughout later chapters generally require absolute convergence.
Problem 1.6.6. Classify \(\displaystyle \int ^{\infty }_{0}\frac {dx}{x^{2/3}(1+x)}\) as an integral of mixed type, split it at a convenient point, and determine whether each part converges.
Problem 1.6.7. The Cauchy principal value of \(\int ^{\infty }_{-\infty }f\) is \(\lim _{R\to \infty }\int ^{R}_{-R}f\). Show that \(\int ^{\infty }_{-\infty }x\,dx\) has principal value \(0\) but does not converge as an improper integral. Explain why the distinction is not pedantic. Where to start: the improper integral requires the two halves to converge separately, with the limits taken independently.
Problem 1.6.8. Show that \(\displaystyle \int ^{\infty }_{0}x^{n}e^{-x}\,dx = n!\) for every non-negative integer \(n\), by induction using integration by parts. This is the result the gamma function of Chapter 2 generalises to non-integer \(n\).
Show solution
Solution. For \(n=0\), \(\int ^{\infty }_{0}e^{-x}dx = 1 = 0!\). Assume the result for \(n-1\). Integrating by parts with \(u=x^{n}\) and \(dv=e^{-x}dx\), \[\int ^{\infty }_{0}x^{n}e^{-x}dx = \left [-x^{n}e^{-x}\right ]^{\infty }_{0} + n\int ^{\infty }_{0}x^{n-1}e^{-x}dx = 0 + n\,(n-1)! = n!,\] the boundary term vanishing at infinity because \(e^{-x}\) decays faster than any power grows, and at zero because \(x^{n}\) vanishes there for \(n\geq 1\).
Problem 1.6.9. For which values of \(s\) does \(\displaystyle \int ^{\infty }_{0}t^{s-1}e^{-t}\,dt\) converge? Treat the two ends separately and state the condition each imposes. Where to start: near zero the exponential is close to \(1\) and the integrand behaves like \(t^{s-1}\); at infinity the exponential dominates any power.
Show solution
Solution. Split at \(t=1\).
Near zero, \(e^{-t}\to 1\), so the integrand behaves like \(t^{s-1}\), and \(\int ^{1}_{0}t^{s-1}dt\) converges precisely when \(s-1>-1\), that is \(s>0\).
At infinity, \(t^{s-1}e^{-t}\leq C e^{-t/2}\) for every \(s\) and \(t\) large enough, since \(t^{s-1}e^{-t/2}\to 0\); so \(\int ^{\infty }_{1}\) converges for every \(s\).
The condition is therefore \(s>0\), and it comes entirely from the behaviour at the origin. This is precisely the domain on which the gamma function is defined by its integral, and it is why extending \(\Gamma \) to \(s\leq 0\) requires the functional equation rather than the integral.
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