The reader is probably familiar with the idea of a logarithm and its use in numerical calculation. He will remember that in elementary algebra logax, the logarithm of x to the base a, is defined by the equations x=ay,y=logax. 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=logx=logex,z=log10x, then x=ey and also x=10z=ezlog10, so that log10x=(logex)/(loge10). Thus it is easy to pass from one system to the other when once loge10 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

Example LXXXVII
1. Show that Dxeaxcosbx=reaxcos(bx+θ),Dxeaxsinbx=reaxsin(bx+θ) where r=a2+b2, cosθ=a/r, sinθ=b/r. Hence determine the nth derivatives of the functions eaxcosbx, eaxsinbx, and show in particular that Dxneax=aneax.

2. Trace the curve y=eaxsinbx, 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=ba2+b2eax.

3. Integrals containing the exponential function. Prove that eaxcosbxdx=acosbx+bsinbxa2+b2eax,eaxsinbxdx=asinbxbcosbxa2+b2eax.

[Denoting the two integrals by I, J, and integrating by parts, we obtain aI=eaxcosbx+bJ,aJ=eaxsinbxbI. Solve these equations for I and J.]

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 ba2+b21+eaπ/b1eaπ/b.

5. Prove that if a>0 then 0eaxcosbxdx=aa2+b2,0eaxsinbxdx=ba2+b2.

6. If In=eaxxndx then aIn=eaxxnnIn1. [Integrate by parts. It follows that In can be calculated for all positive integral values of n.]

7. Prove that, if n is a positive integer, then 0ξexxndx=n!eξ(eξ1ξξ22!ξnn!) and 0exxndx=n!.

8. Show how to find the integral of any rational function of ex. [Put x=logu, when ex=u, dx/du=1/u, and the integral is transformed into that of a rational function of u.]

9. Integrate e2x(c2ex+a2ex)(c2ex+b2ex), 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,eax,ebx,), 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 Axmekx, where m is a positive integer.]

11. Show how to integrate any function of the form P(x, eax, ebx, , coslx, cosmx, , sinlx, sinmx, ).

12. Prove that aeλxR(x)dx, where λ>0 and a is greater than the greatest root of the denominator of R(x), is convergent. [This follows from the fact that eλx tends to infinity more rapidly than any power of x.]

13. Prove that eλx2+μxdx, where λ>0, is convergent for all values of μ, and that the same is true of eλx2+μxxndx, where n is any positive integer.

14. Draw the graphs of ex2, ex2, xex, xex, xex2, xex2, and xlogx, determining any maxima and minima of the functions and any points of inflexion on their graphs.

15. Show that the equation eax=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=eax at the point (ξ,eaξ) is yeaξ=aeaξ(xξ), which passes through the origin if aξ=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 ex=1+x has no real root except x=0, and that ex=1+x+12x2 has three real roots.

17. Draw the graphs of the functions log(x+x2+1),log(1+x1x),eaxcos2bx,e(1/x)2,e(1/x)21/x,ecotx,ecot2x.

18. Determine roughly the positions of the real roots of the equations log(x+x2+1)=x100,ex2+x2x=110,000,exsinx=7,ex2sinx=10,000.

19. The hyperbolic functions. The hyperbolic functions coshx,2 sinhx, … are defined by the equations coshx=12(ex+ex),sinhx=12(exex),tanhx=(sinhx)/(coshx),cothx=(coshx)/(sinhx),sechx=1/(coshx),cosechx=1/(sinhx). Draw the graphs of these functions.

Establish the formulae cosh(x)=coshx,sinh(x)=sinhx,tanh(x)=tanhx,cosh2xsinh2x=1,sech2x+tanh2x=1,coth2xcosech2x=1,cosh2x=cosh2x+sinh2x,sinh2x=2sinhxcoshx,cosh(x+y)=coshxcoshy+sinhxsinhy,sinh(x+y)=sinhxcoshy+coshxsinhy.

21. Verify that these formulae may be deduced from the corresponding formulae in cosx and sinx, by writing coshx for cosx and isinhx for sinx.

[It follows that the same is true of all the formulae involving cosnx and sinnx which are deduced from the corresponding elementary properties of cosx and sinx. The reason of this analogy will appear in Ch. X.]

22. Express coshx and sinhx in terms (a) of cosh2x (b) of sinh2x. Discuss any ambiguities of sign that may occur.

23. Prove that Dxcoshx=sinhx,Dxsinhx=coshx,Dxtanhx=sech2x,Dxcothx=cosech2x,Dxsechx=sechxtanhx,Dxcosechx=cosechxcothx,Dxlogcoshx=tanhx,Dxlog|sinhx|=cothx,Dxarctanex=12sechx,Dxlog|tanh12x|=cosechx.

[All these formulae may of course be transformed into formulae in integration.]

24. Prove that coshx>1 and 1<tanhx<1.

25. Prove that if y=coshx then x=log{y±y21}, if y=sinhx then x=log{y+y2+1}, and if y=tanhx then x=12log{(1+y)/(1y)}. Account for the ambiguity of sign in the first case.

26. We shall denote the functions inverse to coshx, sinhx, tanhx by argcoshx, argsinhx, argtanhx. Show that argcoshx is defined only when x1, and is in general two-valued, while argsinhx is defined for all real values of x, and argtanhx when 1<x<1, and both of the two latter functions are one-valued. Sketch the graphs of the functions.

27. Show that if 12π<x<12π and y is positive, and cosxcoshy=1, then y=log(secx+tanx),Dxy=secx,Dyx=sechy.

28. Prove that if a>0 then dxx2+a2=argsinh(x/a), and dxx2a2 is equal to argcosh(x/a) or to argcosh(x/a), according as x>0 or x<0.

29. Prove that if a>0 then dxx2a2 is equal to (1/a)argtanh(x/a) or to (1/a)argcoth(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 3dx(xa)(xb)=2log{xa+xb}(a<b<x),dx(ax)(bx)=2log{ax+bx}(x<a<b),dx(xa)(bx)=2arctanxabx(a<x<b).

31. Prove that 01xlog(1+12x)dx=3432log32<1201x2dx=16.

32. Solve the equation acoshx+bsinhx=c, where c>0, showing that it has no real roots if b2+c2a2<0, while if b2+c2a2>0 it has two, one, or no real roots according as a+b and ab are both positive, of opposite signs, or both negative. Discuss the case in which b2+c2a2=0.

33. Solve the simultaneous equations coshxcoshy=a, sinhxsinhy=b.

34. x1/x1 as x. [For x1/x=e(logx)/x, and (logx)/x0. Cf. Ex. XXVII. 11.] Show also that the function x1/x has a maximum when x=e, and draw the graph of the function for positive values of x.

35. xx1 as x+0.

36. If {f(n+1)}/{f(n)}l, where l>0, as n, then f(n)nl. [For logf(n+1)logf(n)logl, and so (1/n)logf(n)logl (Ch.IV, Misc. Ex. 27).]

37. n!n/n1/e as n.

[If f(n)=nnn! then {f(n+1)}/{f(n)}={1+(1/n)}n1/e. Now use Ex. 36.]

38. (2n)!/(n!)2n4 as n.

39. Discuss the approximate solution of the equation ex=x1,000,000.

[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 ex=x1,000,000 then x=106logx,logx=13.82+loglogx,loglogx=2.63+log(1+loglogx13.82), roughly, since 13.82 and 2.63 are approximate values of log106 and loglog106 respectively. It is easy to see from these equations that the ratios logx:13.82 and loglogx:2.63 do not differ greatly from unity, and that x=106(13.82+loglogx)=106(13.82+2.63)=16,450,000 gives a tolerable approximation to the root, the error involved being roughly measured by 106(loglogx2.63) or (106loglogx)/13.82 or (106×2.63)/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 ex=1,000,000x1,000,000,ex2=x1,000,000,000.


  1. See for example Chrystal’s Algebra, vol. i, ch. xxi. The value of loge10 is 2.302 and that of its reciprocal .434.↩︎
  2. ‘Hyperbolic cosine’: for an explanation of this phrase see Hobson’s Trigonometry, ch. XVI.↩︎
  3. 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.↩︎

209. The representation of logx as a limit Main Page 211. Logarithmic tests of convergence for series and integrals