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.]