1. Given that \(\log_{10} e = .4343\) and that \(2^{10}\) and \(3^{21}\) are nearly equal to powers of \(10\), calculate \(\log_{10}2\) and \(\log_{10}3\) to four places of decimals.

2. Determine which of \((\frac{1}{2}e)^{\sqrt{3}}\) and \((\sqrt{2})^{\frac{1}{2}\pi}\) is the greater. [Take logarithms and observe that \(\sqrt{3}/(\sqrt{3} + \frac{1}{4}\pi) < \frac{2}{5} \sqrt{3} < .6929 < \log 2\).]

3. Show that \(\log_{10}n\) cannot be a rational number if \(n\) is any positive integer not a power of \(10\). [If \(n\) is not divisible by \(10\), and \(\log_{10}n = p/q\), we have \(10^{p} = n^{q}\), which is impossible, since \(10^{p}\) ends with \(0\) and \(n^{q}\) does not. If \(n = 10^{a}N\), where \(N\) is not divisible by \(10\), then \(\log_{10}N\) and therefore \[\log_{10}n = a + \log_{10}N\] cannot be rational.]

4. For what values of \(x\) are the functions \(\log x\), \(\log\log x\), \(\log\log\log x\), … (a) equal to \(0\) (b) equal to \(1\) (c) not defined? Consider also the same question for the functions \(lx\), \(llx\), \(lllx\), …, where \(lx = \log |x|\).

5. Show that \[\log x – \binom{n}{1} \log(x + 1) + \binom{n}{2} \log(x + 2) – \dots + (-1)^{n} \log(x + n)\] is negative and increases steadily towards \(0\) as \(x\) increases from \(0\) towards \(\infty\).

[The derivative of the function is \[\sum_{0}^{n} (-1)^{r} \binom{n}{r} \frac{1}{x + r} = \frac{n!}{x(x + 1) \dots (x + n)},\] as is easily seen by splitting up the right-hand side into partial fractions. This expression is positive, and the function itself tends to zero as \(x \to \infty\), since \[\log(x + r) = \log x + \epsilon_{x},\] where \(\epsilon_{x} \to 0\), and \(1 – \dbinom{n}{1} + \dbinom{n}{2} – \dots = 0\).]

6. Prove that \[\left(\frac{d}{dx}\right)^{n} \frac{\log x}{x} = \frac{(-1)^{n} n!}{x^{n+1}} \left(\log x – 1 – \frac{1}{2} – \dots – \frac{1}{n}\right).\]

7. If \(x > -1\) then \(x^{2} > (1 + x) \{\log(1 + x)\}^{2}\).

[Put \(1 + x = e^{\xi}\), and use the fact that \(\sinh \xi > \xi\) when \(\xi > 0\).]

8. Show that \(\{\log(1 + x)\}/x\) and \(x/\{(1 + x)\log(1 + x)\}\) both decrease steadily as \(x\) increases from \(0\) towards \(\infty\).

9. Show that, as \(x\) increases from \(-1\) towards \(\infty\), the function \((1 + x)^{-1/x}\) assumes once and only once every value between \(0\) and \(1\).

10. Show that \(\dfrac{1}{\log(1 + x)} – \dfrac{1}{x} \to \dfrac{1}{2}\) as \(x \to 0\).

11. Show that \(\dfrac{1}{\log(1 + x)} – \dfrac{1}{x}\) decreases steadily from \(1\) to \(0\) as \(x\) increases from \(-1\) towards \(\infty\). [The function is undefined when \(x = 0\), but if we attribute to it the value \(\frac{1}{2}\) when \(x = 0\) it becomes continuous for \(x = 0\). Use Ex. 7 to show that the derivative is negative.]

12. Show that the function \((\log \xi – \log x)/(\xi – x)\), where \(\xi\) is positive, decreases steadily as \(x\) increases from \(0\) to \(\xi\), and find its limit as \(x \to \xi\).

13. Show that \(e^{x} > Mx^{N}\), where \(M\) and \(N\) are large positive numbers, if \(x\) is greater than the greater of \(2\log M\) and \(16N^{2}\).

[It is easy to prove that \(\log x < 2\sqrt{x}\); and so the inequality given is certainly satisfied if \[x > \log M + 2N\sqrt{x},\] and therefore certainly satisfied if \(\frac{1}{2}x > \log M\), \(\frac{1}{2}x > 2N\sqrt{x}\).]

14. If \(f(x)\) and \(\phi(x)\) tend to infinity as \(x \to \infty\), and \(f'(x)/\phi'(x) \to \infty\), then \(f(x)/\phi(x) \to \infty\). [Use the result of Ch. VI, Misc. Ex. 33.] By taking \(f(x) = x^{\alpha}\), \(\phi(x) = \log x\), prove that \((\log x)/x^{\alpha} \to 0\) for all positive values of \(\alpha\).

15. If \(p\) and \(q\) are positive integers then \[\frac{1}{pn + 1} + \frac{1}{pn + 2} + \dots + \frac{1}{qn} \to \log\left(\frac{q}{p}\right)\] as \(n \to \infty\). [Cf. Ex. LXXVIII. 6.]

16. Prove that if \(x\) is positive then \(n\log\{\frac{1}{2}(1 + x^{1/n})\} \to -\frac{1}{2}\log x\) as \(n \to \infty\). [We have \[n\log\{\tfrac{1}{2}(1 + x^{1/n})\} = n\log\{1 – \tfrac{1}{2}(1 – x^{1/n})\} = \tfrac{1}{2}n(1 – x^{1/n}) \frac{\log(1 – u)}{u}\] where \(u = \frac{1}{2}(1 – x^{1/n})\). Now use § 209 and Ex. LXXXII. 4.]

17. Prove that if \(a\) and \(b\) are positive then \[\{\tfrac{1}{2}(a^{1/n} + b^{1/n})\}^{n} \to \sqrt{ab}.\]

[Take logarithms and use Ex. 16.]

18. Show that \[1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2n – 1} = \tfrac{1}{2}\log n + \log 2 + \tfrac{1}{2} \gamma + \epsilon_{n},\] where \(\gamma\) is Euler’s constant (Ex. LXXXIX. 1) and \(\epsilon_{n} \to 0\) as \(n \to \infty\).

19. Show that \[1 + \tfrac{1}{3} – \tfrac{1}{2} + \tfrac{1}{5} + \tfrac{1}{7} – \tfrac{1}{4} + \tfrac{1}{9} + \dots = \tfrac{3}{2} \log 2,\] the series being formed from the series \(1 – \frac{1}{2} + \frac{1}{3} – \dots\) by taking alternately two positive terms and then one negative. [The sum of the first \(3n\) terms is \[\begin{gathered} 1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{4n – 1} – \frac{1}{2} \left(1 + \frac{1}{2} + \dots + \frac{1}{n}\right)\\ = \tfrac{1}{2}\log 2n + \log 2 + \tfrac{1}{2}\gamma + \epsilon_{n} – \tfrac{1}{2}(\log n + \gamma + \epsilon_{n}’), \end{gathered}\] where \(\epsilon_{n}\) and \(\epsilon’_{n}\) tend to \(0\) as \(n \to \infty\). (Cf. Ex. LXXVIII. 6).]

20. Show that \(1 – \frac{1}{2} – \frac{1}{4} + \frac{1}{3} – \frac{1}{6} – \frac{1}{8} + \frac{1}{5} – \frac{1}{10} – \dots = \frac{1}{2}\log 2\).

21. Prove that \[\sum_{1}^{n} \frac{1}{\nu(36\nu^{2} – 1)} = -3 + 3\Sigma_{3n+1} – \Sigma_{n} – S_{n}\] where \(S_{n} = 1 + \dfrac{1}{2} + \dots + \dfrac{1}{n}\), \(\Sigma_{n} = 1 + \dfrac{1}{3} + \dots + \dfrac{1}{2n – 1}\). Hence prove that the sum of the series when continued to infinity is \[-3 + \tfrac{3}{2}\log 3 + 2\log 2.\]

22. Show that \[\sum_{1}^{\infty} \frac{1}{n(4n^{2} – 1)} = 2\log 2 – 1, \quad \sum_{1}^{\infty} \frac{1}{n(9n^{2} – 1)} = \tfrac{3}{2}(\log 3 – 1).\]

23. Prove that the sums of the four series \[\sum_{1}^{\infty} \frac{1}{4n^{2} – 1},\quad \sum_{1}^{\infty} \frac{(-1)^{n-1}}{4n^{2} – 1},\quad \sum_{1}^{\infty} \frac{1}{(2n + 1)^{2} – 1},\quad \sum_{1}^{\infty} \frac{(-1)^{n-1}}{(2n + 1)^{2} – 1}\] are \(\frac{1}{2}\), \(\frac{1}{4}\pi – \frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{2}\log 2 – \frac{1}{4}\) respectively.

24. Prove that \(n!\, (a/n)^{n}\) tends to \(0\) or to \(\infty\) according as \(a < e\) or \(a > e\).

[If \(u_{n} = n!\, (a/n)^{n}\) then \(u_{n+1}/u_{n} = a\{1 + (1/n)\}^{-n} \to a/e\). It can be shown that the function tends to \(\infty\) when \(a = e\): for a proof, which is rather beyond the scope of the theorems of this chapter, see Bromwich’s*Infinite Series*, pp. 461

*et seq.*]

25. Find the limit as \(x \to \infty\) of \[\left(\frac{a_{0} + a_{1} x + \dots + a_{r} x^{r}} {b_{0} + b_{1} x + \dots + b_{r} x^{r}}\right)^{\lambda_{0}+\lambda_{1}x},\] distinguishing the different cases which may arise.

26. Prove that \[\sum \log \left(1 + \frac{x}{n}\right)\quad (x > 0)\] diverges to \(\infty\). [Compare with \(\sum (x/n)\).] Deduce that if \(x\) is positive then \[(1 + x)(2 + x) \dots (n + x)/n! \to \infty\] as \(n \to \infty\). [The logarithm of the function is \(\sum\limits_{1}^{n} \log \left(1 + \dfrac{x}{\nu}\right)\).]

27. Prove that if \(x > -1\) then \[\begin{gathered} \frac{1}{(x + 1)^{2}} = \frac{1}{(x + 1) (x + 2)} + \frac{1!}{(x + 1) (x + 2) (x + 3)}\\ + \frac{2!}{(x + 1) (x + 2) (x + 3) (x + 4)} + \dots.\end{gathered}\]

[The difference between \(1/(x + 1)^{2}\) and the sum of the first \(n\) terms of the series is \[\frac{1}{(x + 1)^{2}}\, \frac{n!}{(x + 2) (x + 3) \dots (x + n + 1)}.]\]

28. No equation of the type \[Ae^{\alpha x} + Be^{\beta x} + \dots = 0,\] where \(A\), \(B\), … are polynomials and \(\alpha\), \(\beta\), … different real numbers, can hold for all values of \(x\). [If \(\alpha\) is the algebraically greatest of \(\alpha\), \(\beta\), …, then the term \(Ae^{\alpha x}\) outweighs all the rest as \(x \to \infty\).]

29. Show that the sequence \[a_{1} = e,\quad a_{2} = e^{e^{2}},\quad a_{3} = e^{e^{e^{3}}},\ \dots\] tends to infinity more rapidly than any member of the exponential scale.

[Let \(e_{1}(x) = e^{x}\), \(e_{2}(x) = e^{e_{1}(x)}\), and so on. Then, if \(e_{k}(x)\) is any member of the exponential scale, \(a_{n} > e_{k}(n)\) when \(n > k\).]

30. Prove that \[\frac{d}{dx} \{\phi(x)\}^{\psi(x)} = \frac{d}{dx} \{\phi(x)\}^{\alpha} + \frac{d}{dx} \{\beta^{\psi(x)}\}\] where \(\alpha\) is to be put equal to \(\psi(x)\) and \(\beta\) to \(\phi(x)\) after differentiation. Establish a similar rule for the differentiation of \(\phi(x)^{[\{\psi(x)\}^{\chi(x)}]}\).

31. Prove that if \(D_{x}^{n} e^{-x^{2}} = e^{-x^{2}} \phi_{n}(x)\) then (i) \(\phi_{n}(x)\) is a polynomial of degree \(n\), (ii) \(\phi_{n+1} = -2x\phi_{n} + \phi_{n}’\), and (iii) all the roots of \(\phi_{n} = 0\) are real and distinct, and separated by those of \(\phi_{n-1} = 0\). [To prove (iii) assume the truth of the result for \({\kappa} = 1\), \(2\), …, \({n}\), and consider the signs of \({\phi_{n+1}}\) for the \(n\) values of \(x\) for which \({\phi_{n}} = 0\) and for large (positive or negative) values of \(x\).]

32. The general solution of \(f(xy) = f(x)f(y)\), where \(f\) is a differentiable function, is \(x^{a}\), where \(a\) is a constant: and that of \[f(x + y) + f(x – y) = 2f(x)f(y)\] is \(\cosh ax\) or \(\cos ax\), according as \(f”(0)\) is positive or negative. [In proving the second result assume that \(f\) has derivatives of the first three orders. Then \[2f(x) + y^{2}\{f”(x) + \epsilon_{y}\} = 2f(x)[f(0) + yf'(0) + \tfrac{1}{2} y^{2}\{f”(0) + \epsilon_{y}’\}],\] where \(\epsilon_{y}\) and \(\epsilon_{y}’\) tend to zero with \(y\). It follows that \(f(0) = 1\), \(f'(0) = 0\), \(f”(x) = f”(0)f(x)\), so that \(a = \sqrt{f”(0)}\) or \(a = \sqrt{-f”(0)}\).]

33. How do the functions \(x^{\sin(1/x)}\), \(x^{\sin^{2}(1/x)}\), \(x^{\csc(1/x)}\) behave as \(x \to +0\)?

34. Trace the curves \(y = \tan x e^{\tan x}\), \(y = \sin x \log \tan \frac{1}{2}x\).

35. The equation \(e^{x} = ax + b\) has one real root if \(a < 0\) or \(a = 0\), \(b > 0\). If \(a > 0\) then it has two real roots or none, according as \(a\log a > b – a\) or \(a\log a < b – a\).

36. Show by graphical considerations that the equation \(e^{x} = ax^{2} + 2bx + c\) has one, two, or three real roots if \(a > 0\), none, one, or two if \(a < 0\); and show how to distinguish between the different cases.

37. Trace the curve \(y = \dfrac{1}{x} \log\left(\dfrac{e^{x} – 1}{x}\right)\), showing that the point \((0, \frac{1}{2})\) is a centre of symmetry, and that as \(x\) increases through all real values, \(y\) steadily increases from \(0\) to \(1\). Deduce that the equation \[\frac{1}{x} \log\left(\frac{e^{x} – 1}{x}\right) = \alpha\] has no real root unless \(0 < \alpha < 1\), and then one, whose sign is the same as that of \(\alpha – \frac{1}{2}\). [In the first place \[y – \tfrac{1}{2} = \frac{1}{x} \left\{\log\left(\frac{e^{x} – 1}{x}\right) – \log e^{\frac{1}{2} x}\right\} = \frac{1}{x} \log\left(\frac{\sinh \frac{1}{2}x}{\frac{1}{2}x}\right)\] is clearly an odd function of \(x\). Also \[\frac{dy}{dx} = \frac{1}{x^{2}} \left\{\tfrac{1}{2} x\coth \tfrac{1}{2}x – 1 – \log\left(\frac{\sinh \frac{1}{2}x}{\frac{1}{2}x}\right)\right\}.\] The function inside the large bracket tends to zero as \(x \to 0\); and its derivative is \[\frac{1}{x} \left\{1 – \left(\frac{\frac{1}{2}x}{\sinh \frac{1}{2}x}\right)^2\right\},\] which has the sign of \(x\). Hence \(dy/dx > 0\) for all values of \(x\).]

38. Trace the curve \(y = e^{1/x} \sqrt{x^{2} + 2x}\), and show that the equation \[e^{1/x} \sqrt{x^{2} + 2x} = \alpha\] has no real roots if \(\alpha\) is negative, one negative root if \[0 < \alpha < a = e^{1/\sqrt{2}} \sqrt{2 + 2\sqrt{2}},\] and two positive roots and one negative if \(\alpha > a\).

39. Show that the equation \(f_{n}(x) = 1 + x + \dfrac{x^{2}}{2!} + \dots + \dfrac{x^{n}}{n!} = 0\) has one real root if \(n\) is odd and none if \(n\) is even.

[Assume this proved for \(n = 1\), \(2\), … \(2k\). Then \(f_{2k+1}(x) = 0\) has at least one real root, since its degree is odd, and it cannot have more since, if it had, \(f’_{2k+1}(x)\) or \(f_{2k}(x)\) would have to vanish once at least. Hence \(f_{2k+1}(x) = 0\) has just one root, and so \(f_{2k+2}(x) = 0\) cannot have more than two. If it has two, say \(\alpha\) and \(\beta\), then \(f’_{2k+2}(x)\) or \(f_{2k+1}(x)\) must vanish once at least between \(\alpha\) and \(\beta\), say at \(\gamma\). And \[f_{2k+2}(\gamma) = f_{2k+1}(\gamma) + \frac{\gamma^{2k+2}}{(2k + 2)!} > 0.\] But \(f_{2k+2}(x)\) is also positive when \(x\) is large (positively or negatively), and a glance at a figure will show that these results are contradictory. Hence \(f_{2k+2}(x) = 0\) has no real roots.]

40. Prove that if \(a\) and \(b\) are positive and nearly equal then \[\log \frac{a}{b} = \frac{1}{2}(a – b) \left(\frac{1}{a} + \frac{1}{b}\right),\] approximately, the error being about \(\frac{1}{6}\{(a – b)/a\}^{3}\). [Use the logarithmic series. This formula is interesting historically as having been employed by Napier for the numerical calculation of logarithms.]

41. Prove by multiplication of series that if \(-1 < x < 1\) then \[\begin{aligned} \tfrac{1}{2}\{\log(1 + x)\}^{2} &= \tfrac{1}{2} x^{2} – \tfrac{1}{3}(1 + \tfrac{1}{2})x^{3} + \tfrac{1}{4}(1 + \tfrac{1}{2} + \tfrac{1}{3})x^{4} – \dots,\\ \tfrac{1}{2}(\arctan x)^{2} &= \tfrac{1}{2} x^{2} – \tfrac{1}{4}(1 + \tfrac{1}{3})x^{4} + \tfrac{1}{6}(1 + \tfrac{1}{3} + \tfrac{1}{5})x^{6} – \dots.\end{aligned}\]

42. Prove that \[(1 + \alpha x)^{1/x} = e^{\alpha}\{1 – \tfrac{1}{2} a^{2}x + \tfrac{1}{24}(8 + 3a)a^{3}x^{2}(1 + \epsilon_{x})\},\] where \(\epsilon_{x} \to 0\) with \(x\).

43. The first \(n + 2\) terms in the expansion of \(\log\left(1 + x + \dfrac{x^{2}}{2!} + \dots + \dfrac{x^{n}}{n!}\right)\) in powers of \(x\) are \[x – \frac{x^{n+1}}{n!} \left\{\frac{1}{n + 1} – \frac{x}{1!\, (n + 2)} + \frac{x^{2}}{2!\, (n + 3)} – \dots + (-1)^{n} \frac{x^{n}}{n!\, (2n + 1)} \right\}.\]

44. Show that the expansion of \[\exp \left(-x – \frac{x^{2}}{2} – \dots – \frac{x^{n}}{n}\right)\] in powers of \(x\) begins with the terms \[1 – x + \frac{x^{n+1}}{n + 1} – \sum_{s=1}^{n} \frac{x^{n+s+1}}{(n + s)(n + s + 1)}.\]

45. Show that if \(-1 < x < 1\) then \[\begin{aligned} \frac{1}{3}x + \frac{1\cdot4}{3\cdot6}2^{2}x^{2} + \frac{1\cdot4\cdot7}{3\cdot6\cdot9}3^{2}x^{3} + \dots &= \frac{x(x + 3)}{9(1 – x)^{7/3}},\\ \frac{1}{3}x + \frac{1\cdot4}{3\cdot6}2^{3}x^{2} + \frac{1\cdot4\cdot7}{3\cdot6\cdot9}3^{3}x^{3} + \dots &= \frac{x(x^{2} + 18x + 9)}{27(1 – x)^{10/3}}.\end{aligned}\]

[Use the method of Ex. XCII. 6. The results are more easily obtained by differentiation; but the problem of the differentiation of an infinite series is beyond our range.]

46. Prove that \[\begin{aligned} \int_{0}^{\infty} \frac{dx}{(x + a)(x + b)} &= \frac{1}{a – b} \log\left(\frac{a}{b}\right), \\ \int_{0}^{\infty} \frac{dx}{(x + a)(x + b)^{2}} &= \frac{1}{(a – b)^{2}b}\left\{a – b – b\log\left(\frac{a}{b}\right)\right\},\\ \int_{0}^{\infty} \frac{x\, dx}{(x + a)(x + b)^{2}} &= \frac{1}{(a – b)^{2}} \left\{a\log\left(\frac{a}{b}\right) – a + b\right\},\\ \int_{0}^{\infty} \frac{dx}{(x + a)(x^{2} + b^{2})} &= \frac{1}{(a^{2} + b^{2})b} \left\{\tfrac{1}{2}\pi a – b\log\left(\frac{a}{b}\right)\right\},\\ \int_{0}^{\infty} \frac{x\, dx}{(x + a)(x^{2} + b^{2})} &= \frac{1}{a^{2} + b^{2}} \left\{\tfrac{1}{2}\pi b + a\log\left(\frac{a}{b}\right)\right\},\end{aligned}\] provided that \(a\) and \(b\) are positive. Deduce, and verify independently, that each of the functions \[a – 1 – \log a,\quad a\log a – a + 1,\quad \tfrac{1}{2}\pi a – \log a,\quad \tfrac{1}{2}\pi + a\log a\] is positive for all positive values of \(a\).

47. Prove that if \(\alpha\), \(\beta\), \(\gamma\) are all positive, and \(\beta^{2} > \alpha\gamma\), then \[\int_{0}^{\infty} \frac{dx}{\alpha x^{2} + 2\beta x + \gamma} = \frac{1}{\sqrt{\beta^{2} – \alpha\gamma}} \log \left\{\frac{\beta + \sqrt{\beta^{2} – \alpha\gamma}} {\sqrt{\alpha\gamma}} \right\};\] while if \(\alpha\) is positive and \(\alpha\gamma > \beta^{2}\) the value of the integral is \[\frac{1}{\sqrt{\alpha\gamma – \beta^{2}}} \arctan \left\{\frac{\sqrt{\alpha\gamma – \beta^{2}}}{\beta}\right\},\] that value of the inverse tangent being chosen which lies between \(0\) and \(\pi\). Are there any other really different cases in which the integral is convergent?

48. Prove that if \(a > -1\) then \[\int_{1}^{\infty} \frac{dx}{(x + a)\sqrt{x^{2} – 1}} = \int_{0}^{\infty} \frac{dt}{\cosh t + a} = 2\int_{1}^{\infty}\frac{du}{u^{2} + 2au + 1};\] and deduce that the value of the integral is \[\frac{2}{\sqrt{1 – a^{2}}} \arctan \sqrt{\frac{1 – a}{1 + a}}\] if \(-1 < a < 1\), and \[\frac{1}{\sqrt{a^{2} – 1}} \log\frac{\sqrt{a + 1} + \sqrt{a – 1}} {\sqrt{a + 1} – \sqrt{a – 1}} = \frac{2}{\sqrt{a^{2} – 1}} \operatorname{arg tanh} \sqrt{\frac{a – 1}{a + 1}}\] if \(a > 1\). Discuss the case in which \(a = 1\).

49. Transform the integral \(\int_{0}^{\infty} \frac{dx}{(x + a) \sqrt{x^{2} + 1}}\), where \(a > 0\), in the same ways, showing that its value is \[\frac{1}{\sqrt{a^{2} + 1}} \log\frac{a + 1 + \sqrt{a^{2} + 1}}{a + 1 – \sqrt{a^{2} + 1}} = \frac{2}{\sqrt{a^{2} + 1}} \operatorname{arg tanh} \frac{\sqrt{a^{2} + 1}}{a + 1}.\]

50. Prove that \[\int_{0}^{1} \arctan x\, dx = \tfrac{1}{4}\pi – \tfrac{1}{2}\log 2.\]

51. If \(0 < \alpha < 1\), \(0 < \beta < 1\), then \[\int_{-1}^{1} \frac{dx}{\sqrt{(1 – 2\alpha x + \alpha^{2})(1 – 2\beta x + \beta^{2})}} = \frac{1}{\sqrt{\alpha\beta}} \log \frac{1 + \sqrt{\alpha\beta}}{1 – \sqrt{\alpha\beta}}.\]

52. Prove that if \(a > b > 0\) then \[\int_{-\infty}^{\infty} \frac{d\theta}{a\cosh \theta + b\sinh \theta} = \frac{\pi}{\sqrt{a^{2} – b^{2}}}{.}\]

53. Prove that \[\int_{0}^{1} \frac{\log x}{1 + x^{2}}\, dx = -\int_{1}^{\infty} \frac{\log x}{1 + x^{2}}\, dx,\quad \int_{0}^{\infty} \frac{\log x}{1 + x^{2}}\, dx = 0,\] and deduce that if \(a > 0\) then \[\int_{0}^{\infty} \frac{\log x}{a^{2} + x^{2}}\, dx = \frac{\pi}{2a}\log a.\]

[Use the substitutions \(x = 1/t\) and \(x = au\).]54. Prove that \[\int_{0}^{\infty} \log \left(1 + \frac{a^{2}}{x^{2}}\right) dx = \pi a\] if \(a > 0\). [Integrate by parts.]

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