6 integrals from to if the integrals z a φ x dx z a

Info icon This preview shows pages 419–422. Sign up to view the full content.

View Full Document Right Arrow Icon
6. Integrals from -∞ to + . If the integrals Z a -∞ φ ( x ) dx, Z a φ ( x ) dx are both convergent, and have the values k , l respectively, then we say that Z -∞ φ ( x ) dx is convergent and has the value k + l . 7. Prove that Z 0 -∞ dx 1 + x 2 = Z 0 dx 1 + x 2 = 1 2 Z -∞ dx 1 + x 2 = 1 2 π. 8. Prove generally that Z -∞ φ ( x 2 ) dx = 2 Z 0 φ ( x 2 ) dx, provided that the integral Z 0 φ ( x 2 ) dx is convergent. 9. Prove that if Z 0 ( x 2 ) dx is convergent then Z -∞ ( x 2 ) dx = 0. 10. Analogue of Abel’s Theorem of § 173 . If φ ( x ) is positive and steadily decreases, and Z a φ ( x ) dx is convergent, then ( x ) 0 . Prove this ( a ) by means of Abel’s Theorem and the Integral Test and ( b ) directly, by arguments analogous to those of § 173 . 11. If a = x 0 < x 1 < x 2 < . . . and x n → ∞ , and u n = Z x n +1 x n φ ( x ) dx , then the convergence of Z a φ ( x ) dx involves that of u n . If φ ( x ) is always positive the converse statement is also true. [That the converse is not true in general is shown by the example in which φ ( x ) = cos x , x n = .]
Image of page 419

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
[VIII : 179] THE CONVERGENCE OF INFINITE SERIES, ETC. 404 179. Application to infinite integrals of the rules for substitu- tion and integration by parts. The rules for the transformation of a definite integral which were discussed in § 161 may be extended so as to apply to infinite integrals. (1) Transformation by substitution. Suppose that Z a φ ( x ) dx (1) is convergent. Further suppose that, for any value of ξ greater than a , we have, as in § 161 , Z ξ a φ ( x ) dx = Z τ b φ { f ( t ) } f 0 ( t ) dt, (2) where a = f ( b ), ξ = f ( τ ). Finally suppose that the functional relation x = f ( t ) is such that x → ∞ as t → ∞ . Then, making τ and so ξ tend to in (2), we see that the integral Z b φ { f ( t ) } f 0 ( t ) dt (3) is convergent and equal to the integral (1). On the other hand it may happen that ξ → ∞ as τ → -∞ or as τ c . In the first case we obtain Z a φ ( x ) dx = lim τ →-∞ Z τ b φ { f ( t ) } f 0 ( t ) dt = - lim τ →-∞ Z b τ φ { f ( t ) } f 0 ( t ) dt = - Z b -∞ φ { f ( t ) } f 0 ( t ) dt. In the second case we obtain Z a φ ( x ) dx = lim τ c Z τ b φ { f ( t ) } f 0 ( t ) dt. (4) We shall return to this equation in § 181 .
Image of page 420
[VIII : 179] THE CONVERGENCE OF INFINITE SERIES, ETC. 405 There are of course corresponding results for the integrals Z a -∞ φ ( x ) dx, Z -∞ φ ( x ) dx, which it is not worth while to set out in detail: the reader will be able to formulate them for himself. Examples LXXIV. 1. Show, by means of the substitution x = t α , that if s > 1 and α > 0 then Z 1 x - s dx = α Z 1 t α (1 - s ) - 1 dt ; and verify the result by calculating the value of each integral directly. 2. If Z a φ ( x ) dx is convergent then it is equal to one or other of α Z ( a - β ) φ ( αt + β ) dt, - α Z ( a - β ) -∞ φ ( αt + β ) dt, according as α is positive or negative. 3. If φ ( x ) is a positive and steadily decreasing function of x , and α and β are any positive numbers, then the convergence of the series φ ( n ) implies and is implied by that of the series φ ( αn + β ).
Image of page 421

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 422
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern