The reader is probably familiar with the idea of a logarithm and its use in numerical calculation. He will remember that in elementary algebra \(\log_{a} x\), the logarithm of \(x\) to the base \(a\), is defined by the equations \[x = a^{y},\quad y = \log_{a} x.\] This definition is of course applicable only when \(y\) is rational, though this point is often passed over in silence.

Our logarithms are therefore logarithms to the base \(e\). For numerical work logarithms to the base \(10\) are used. If \[y = \log x = \log_{e} x,\quad z = \log_{10} x,\] then \(x = e^{y}\) and also \(x = 10^{z} = e^{z\log 10}\), so that \[\log_{10} x = (\log_{e} x)/(\log_{e} 10).\] Thus it is easy to pass from one system to the other when once \(\log_{e} 10\) has been calculated.

It is no part of our purpose in this book to go into details concerning the practical uses of logarithms. If the reader is not familiar with them he should consult some text-book on Elementary Algebra or Trigonometry.^{1}

2. Trace the curve \(y = e^{-ax}\sin bx\), where \(a\) and \(b\) are positive. Show that \(y\) has an infinity of maxima whose values form a geometrical progression and which lie on the curve \[y = \frac{b}{\sqrt{a^{2} + b^{2}}}\, e^{-ax}.\]

3. **Integrals containing the exponential function.** Prove that \[\begin{aligned} \int e^{ax}\cos bx\, dx &= \frac{a\cos bx + b\sin bx}{a^{2} + b^{2}}\, e^{ax}, \\ \int e^{ax}\sin bx\, dx &= \frac{a\sin bx – b\cos bx}{a^{2} + b^{2}}\, e^{ax}.\end{aligned}\]

4. Prove that the successive areas bounded by the curve of Ex. 2 and the positive half of the axis of \(x\) form a geometrical progression, and that their sum is \[\frac{b}{a^{2} + b^{2}}\, \frac{1 + e^{-a\pi/b}}{1 – e^{{-}a\pi/b}}.\]

5. Prove that if \(a > 0\) then \[\int_{0}^{\infty} e^{-ax}\cos bx\, dx = \frac{a}{a^{2} + b^{2}},\quad \int_{0}^{\infty} e^{-ax}\sin bx\, dx = \frac{b}{a^{2} + b^{2}}.\]

6. If \(I_{n} = \int e^{ax}x^{n}\, dx\) then \(aI_{n} = e^{ax}x^{n} – nI_{n-1}\). [Integrate by parts. It follows that \(I_{n}\) can be calculated for all positive integral values of \(n\).]

7. Prove that, if \(n\) is a positive integer, then \[\int_{0}^{\xi} e^{-x}x^{n}\, dx = n!\, e^{-\xi} \left( e^{\xi} – 1 – \xi – \frac{\xi^{2}}{2!} – \dots – \frac{\xi^{n}}{n!} \right)\] and \[\int_{0}^{\infty} e^{-x}x^{n}\, dx = n!.\]

8. Show how to find the integral of any rational function of \(e^{x}\). [Put \(x = \log u\), when \(e^{x} = u\), \(dx/du = 1/u\), and the integral is transformed into that of a rational function of \(u\).]

9. Integrate \[\frac{e^{2x}}{(c^{2}e^{x} + a^{2}e^{-x})(c^{2}e^{x} + b^{2}e^{-x})},\] distinguishing the cases in which \(a\) is and is not equal to \(b\).

10. Prove that we can integrate any function of the form \(P(x, e^{ax}, e^{bx}, \dots)\), where \(P\) denotes a polynomial. [This follows from the fact that \(P\) can be expressed as the sum of a number of terms of the type \(Ax^{m}e^{kx}\), where \(m\) is a positive integer.]

11. Show how to integrate any function of the form \[P(x,\ e^{ax},\ e^{bx},\ \dots,\ \cos lx,\ \cos mx,\ \dots,\ \sin lx,\ \sin mx,\ \dots).\]

12. Prove that \(\int_{a}^{\infty} e^{-\lambda x} R(x)\, dx\), where \(\lambda > 0\) and \(a\) is greater than the greatest root of the denominator of \(R(x)\), is convergent. [This follows from the fact that \(e^{\lambda x}\) tends to infinity more rapidly than any power of \(x\).]

13. Prove that \(\int_{-\infty}^{\infty} e^{-\lambda x^{2} + \mu x}\, dx\), where \(\lambda > 0\), is convergent for all values of \(\mu\), and that the same is true of \(\int_{-\infty}^{\infty} e^{-\lambda x^{2} + \mu x} x^{n}\, dx\), where \(n\) is any positive integer.

14. Draw the graphs of \(e^{x^{2}}\), \(e^{-x^{2}}\), \(xe^{x}\), \(xe^{-x}\), \(xe^{x^{2}}\), \(xe^{-x^{2}}\), and \(x\log x\), determining any maxima and minima of the functions and any points of inflexion on their graphs.

15. Show that the equation \(e^{ax} = bx\), where \(a\) and \(b\) are positive, has two real roots, one, or none, according as \(b > ae\), \(b = ae\), or \(b < ae\). [The tangent to the curve \(y = e^{ax}\) at the point \((\xi, e^{a\xi})\) is \[y – e^{a\xi} = ae^{a\xi}(x – \xi),\] which passes through the origin if \(a\xi = 1\), so that the line \(y = aex\) touches the curve at the point \((1/a, e)\). The result now becomes obvious when we draw the line \(y = bx\). The reader should discuss the cases in which \(a\) or \(b\) or both are negative.]

16. Show that the equation \(e^{x} = 1 + x\) has no real root except \(x = 0\), and that \(e^{x} = 1 + x + \frac{1}{2}x^{2}\) has three real roots.

17. Draw the graphs of the functions \[\begin{gathered} \log(x + \sqrt{x^{2} + 1}),\quad \log\left(\frac{1 + x}{1 – x}\right),\quad e^{-ax}\cos^{2}bx,\\ e^{-(1/x)^{2}},\quad e^{-(1/x)^{2}}\sqrt{1/x},\quad e^{-\cot x},\quad e^{-\cot^{2} x}.\end{gathered}\]

18. Determine roughly the positions of the real roots of the equations \[\log(x + \sqrt{x^{2} + 1}) = \frac{x}{100},\quad e^{x} – \frac{2 + x}{2 – x} = \frac{1}{10,000},\quad e^{x}\sin x = 7,\quad e^{x^{2}}\sin x = 10,000.\]

19.** The hyperbolic functions.** The hyperbolic functions \(\cosh x\),^{2} \(\sinh x\), … are defined by the equations \[\begin{gathered} \cosh x = \tfrac{1}{2}(e^{x} + e^{-x}),\quad \sinh x = \tfrac{1}{2}(e^{x} – e^{-x}), \\ \tanh x = (\sinh x)/(\cosh x),\quad \coth x = (\cosh x)/(\sinh x),\\ \operatorname{sech} x = 1/(\cosh x),\quad \operatorname{cosech} x = 1/(\sinh x).\end{gathered}\] Draw the graphs of these functions.

Establish the formulae \[\begin{gathered} \cosh(-x) = \cosh x,\quad \sinh(-x) = -\sinh x,\quad \tanh(-x) = -\tanh x,\\ \cosh^{2} x – \sinh^{2} x = 1,\quad \operatorname{sech}^{2} x + \tanh^{2} x = 1,\quad \coth^{2} x – \operatorname{cosech}^{2} x = 1, \\ \cosh 2x = \cosh^{2} x + \sinh^{2} x,\quad \sinh 2x = 2\sinh x\cosh x, \\ \begin{alignedat}{2} \cosh(x + y) &= \cosh x\cosh y &&+ \sinh x\sinh y,\\ \sinh(x + y) &= \sinh x\cosh y &&+ \cosh x\sinh y. \end{alignedat}\end{gathered}\]

21. Verify that these formulae may be deduced from the corresponding formulae in \(\cos x\) and \(\sin x\), by writing \(\cosh x\) for \(\cos x\) and \(i\sinh x\) for \(\sin x\).

[It follows that the same is true of all the formulae involving \(\cos nx\) and \(\sin nx\) which are deduced from the corresponding elementary properties of \(\cos x\) and \(\sin x\). The reason of this analogy will appear in Ch. X.]22. Express \(\cosh x\) and \(\sinh x\) in terms (a) of \(\cosh 2x\) (b) of \(\sinh 2x\). Discuss any ambiguities of sign that may occur.

23. Prove that \[\begin{gathered} D_{x}\cosh x = \sinh x,\quad D_{x}\sinh x = \cosh x,\\ D_{x}\tanh x = \operatorname{sech}^{2}x,\quad D_{x}\coth x = -\operatorname{cosech}^{2}x,\\ D_{x}\operatorname{sech} x = -\operatorname{sech} x\tanh x,\quad D_{x}\operatorname{cosech} x = -\operatorname{cosech} x\coth x,\\ D_{x}\log \cosh x = \tanh x,\quad D_{x}\log|\sinh x| = \coth x,\\ D_{x}\arctan e^{x} = \tfrac{1}{2}\operatorname{sech} x,\quad D_{x}\log |\tanh \tfrac{1}{2} x| = \operatorname{cosech} x.\end{gathered}\]

[All these formulae may of course be transformed into formulae in integration.]24. Prove that \(\cosh x > 1\) and \(-1 < \tanh x < 1\).

25. Prove that if \(y = \cosh x\) then \(x = \log\{y \pm \sqrt{y^{2} – 1}\}\), if \(y = \sinh x\) then \(x = \log\{y + \sqrt{y^{2} + 1}\}\), and if \(y = \tanh x\) then \(x = \frac{1}{2}\log\{(1 + y)/(1 – y)\}\). Account for the ambiguity of sign in the first case.

26. We shall denote the functions inverse to \(\cosh x\), \(\sinh x\), \(\tanh x\) by \(\operatorname{arg cosh} x\), \(\operatorname{arg sinh} x\), \(\operatorname{arg tanh} x\). Show that \(\operatorname{arg cosh} x\) is defined only when \(x \geq 1\), and is in general two-valued, while \(\operatorname{arg sinh} x\) is defined for all real values of \(x\), and \(\operatorname{arg tanh} x\) when \(-1 < x < 1\), and both of the two latter functions are one-valued. Sketch the graphs of the functions.

27. Show that if \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\) and \(y\) is positive, and \(\cos x\cosh y = 1\), then \[y = \log(\sec x + \tan x),\quad D_{x} y = \sec x,\quad D_{y} x = \operatorname{sech} y.\]

28. Prove that if \(a > 0\) then \(\int \frac{dx}{\sqrt{x^{2} + a^{2}}} = \operatorname{arg sinh}(x/a)\), and \(\int \frac{dx}{\sqrt{x^{2} – a^{2}}}\) is equal to \(\operatorname{arg cosh}(x/a)\) or to \(-\operatorname{arg cosh}(-x/a)\), according as \(x > 0\) or \(x < 0\).

29. Prove that if \(a > 0\) then \(\int \frac{dx}{x^{2} – a^{2}}\) is equal to \(-(1/a)\operatorname{arg tanh}(x/a)\) or to \(-(1/a)\operatorname{arg coth}(x/a)\), according as \(|x|\) is less than or greater than \(a\). [The results of Exs. 28 and 29 furnish us with an alternative method of writing a good many of the formulae of Ch.VI.]

30. Prove that \[\begin{aligned} {3} \int \frac{dx}{\sqrt{(x – a)(x – b)}} &= &&2\log\{\sqrt{x – a} + \sqrt{x – b}\} &&(a < b < x),\\ \int \frac{dx}{\sqrt{(a – x)(b – x)}} &= -&&2\log\{\sqrt{a – x} + \sqrt{b – x}\}\quad &&(x < a < b),\\ \int \frac{dx}{\sqrt{(x – a)(b – x)}} &= &&2\arctan\sqrt{\frac{x – a}{b – x}} &&(a < x < b).\end{aligned}\]

31. Prove that \[\int_{0}^{1} x \log(1 + \tfrac{1}{2}x)\, dx = \tfrac{3}{4} – \tfrac{3}{2}\log\tfrac{3}{2} < \tfrac{1}{2} \int_{0}^{1} x^{2}\, dx = \tfrac{1}{6}.\]

32. Solve the equation \(a\cosh x + b\sinh x = c\), where \(c > 0\), showing that it has no real roots if \(b^{2} + c^{2} – a^{2} < 0\), while if \(b^{2} + c^{2} – a^{2} > 0\) it has two, one, or no real roots according as \(a + b\) and \(a – b\) are both positive, of opposite signs, or both negative. Discuss the case in which \(b^{2} + c^{2} – a^{2} = 0\).

33. Solve the simultaneous equations \(\cosh x\cosh y = a\), \(\sinh x\sinh y = b\).

34. \(x^{1/x} \to 1\) as \(x \to \infty\). [For \(x^{1/x} = e^{(\log x)/x}\), and \((\log x)/x \to 0\). Cf. Ex. XXVII. 11.] Show also that the function \(x^{1/x}\) has a maximum when \(x = e\), and draw the graph of the function for positive values of \(x\).

35. \(x^{x} \to 1\) as \(x \to +0\).

36. If \(\{f(n + 1)\}/\{f(n)\} \to l\), where \(l > 0\), as \(n \to \infty\), then \(\sqrt[n]{f(n)} \to l\). [For \(\log f(n + 1) – \log f(n) \to \log l\), and so \((1/n)\log f(n) \to \log l\) (Ch.IV, Misc. Ex. 27).]

37. \(\sqrt[n]{n!}/n \to 1/e\) as \(n \to \infty\).

[If \(f(n) = n^{-n} n!\) then \(\{f(n + 1)\}/\{f(n)\} = \{1 + (1/n)\}^{-n} \to 1/e\). Now use Ex. 36.]38. \(\sqrt[n]{(2n)!/(n!)^{2}} \to 4\) as \(n \to \infty\).

39. Discuss the approximate solution of the equation \(e^{x} = x^{1,000,000}\).

[It is easy to see by general graphical considerations that the equation has two positive roots, one a little greater than \(1\) and one very large,^{3}and one negative root a little greater than \(-1\). To determine roughly the size of the large positive root we may proceed as follows. If \(e^{x} = x^{1,000,000}\) then \[x = 10^{6} \log x,\quad \log x = 13.82 + \log\log x,\quad \log\log x = 2.63 + \log \left(1 + \frac{\log\log x}{13.82}\right),\] roughly, since \(13.82\) and \(2.63\) are approximate values of \(\log 10^{6}\) and \(\log\log 10^{6}\) respectively. It is easy to see from these equations that the ratios \(\log x : 13.82\) and \(\log\log x : 2.63\) do not differ greatly from unity, and that \[x = 10^{6}(13.82 + \log\log x) = 10^{6}(13.82 + 2.63) = 16,450,000\] gives a tolerable approximation to the root, the error involved being roughly measured by \(10^{6}(\log\log x – 2.63)\) or \((10^{6} \log\log x)/13.82\) or \((10^{6} \times 2.63)/13.82\), which is less than \(200,000\). The approximations are of course very rough, but suffice to give us a good idea of the scale of magnitude of the root.]

40. Discuss similarly the equations \[e^{x} = 1,000,000 x^{1,000,000},\quad e^{x^{2}} = x^{1,000,000,000}.\]

- See for example Chrystal’s
*Algebra*, vol. i, ch. xxi. The value of \(\log_{e} 10\) is \(2.302\dots\) and that of its reciprocal \(.434\dots\).↩︎ - ‘Hyperbolic cosine’: for an explanation of this phrase see Hobson’s
*Trigonometry*, ch. XVI.↩︎ - The phrase ‘very large’ is of course not used here in the technical sense explained in Ch.IV. It means ‘a good deal larger than the roots of such equations as usually occur in elementary mathematics’. The phrase ‘a little greater than’ must be interpreted similarly.↩︎

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