Special Applications

<span class='text_page_counter'>(1)</span>SPECIAL APPLICATIONS ADITYA NARAYAN SHARMA - KANCHRAPARA - INDIA. 1. Show that Z. ∞. p. e−x − e−x. q. dx =. p−q. γ x pq for p, q > 0, where γ is the Euler - Mascheroni constant. 0. Proof. Z. ∞. p. q. (e−x − e−x )d(ln x). I= 0. Z h i∞ p q +p I = (e−x − e−x ) ln 0. ∞. p. e−x xp−1 ln x − q. 0. ∞. Z. q. e−x xq−1 ln xdx. 0. The substitutions xp = u, xq = v makes the last two integrals as, Z ∞ Z ∞ Z ∞ q − p −xp p−1 −xq q−1 p e x ln x − q e x ln xdx = e−x x ln xdx qp 0 0 Z ∞ 0 q − p p − q = e−x ln xdx =γ qp qp 0 h i ∞ p q ln x ln x − lim Now (e−x − e−x ) ln x = lim 1 1 0. x→∞. p. e−x −e−x. q. x→0. p. q. e−x −e−x. Both of these can be evaluated by applying L-Hospital’s rule successively and both of them are 0 Z ∞ −xp q p − q (e − e−x ) Hence dx = γ x pq 0  2. Z. 1. tα−1 − tβ−1. 0. (1 + t) ln t. dx = ln. Γ( 12 + 12 α)Γ( 21 β). !. Γ( 12 + 12 β)Γ( 21 α). Proof. 1. xa−1 dx then by differentiating w.r.t to a 0 (1 + x) ln x Z 1 a−1 X x 1 0 F (a) = dx = (−1)n 1 + x n + a 0 n≥0  1 a + 1  a  X X 1 1 1 X 1 = Now, (−1)n = ψ −ψ a − a+1 n+a 2 n+ 2 2 2 2 n+ 2 n≥0 n≥0 n≥0           1 α+1 α 1 β+1 β So, F 0 (α) − F 0 (β) = ψ −ψ − ψ −ψ 2 2 2 2 2 2 Z. Let F (a) =. 1.

<span class='text_page_counter'>(2)</span> 2. ADITYA NARAYAN SHARMA - KANCHRAPARA - INDIA. !  Γ( 1+α )Γ( β )  2 2 +C Integrating we have, F (α) − F (β) = ln α Γ( 1+β 2 )Γ( 2 ) Now since, F (1) = 0 ⇒ C = 0 ! β  Γ( 1+α) 2 )Γ( 2 ) So F (α) − F (β) = ln α Γ( 1+β 2 )Γ( 2 )  3. Find a closed form for Z. ∞. 0. sin2n+1 x. x for n ≥ 0. dx. Proof. Z ∞Z ∞ sin2n+1 x I= e−xy sin2n+1 dxdy dx = x 0 0 0 Now by IBP two times we can create a recurrence relation and thus evaluate, Z ∞ (2n + 1)! Jn = e−ax sin2n+1 xdx = Qn 2 + (2r + 1)2 ) (a 0 r=1 Z ∞ dy Qn Thus, I = (2n + 1)! 2 2 0 r=1 (a + (2r + 1) ) n X Ar 1 = The expression Qn 2 + (2r + 1)2 2 2 y (a + (2r + 1) ) r=1 r=0 Z. ∞. where the coefficients can be determined by the cover-up rule as (−1)k 2k + 1 Ak = 2n · 2 (n − k)!(n + k + 1)!     n Z ∞ n X Ak π X π 2n k 2n + 1 So, I = (2n+1)! dy = (−1) = 2 + (2k + 1)2 2n+1 2n+1 n − k n y 2 2 0 k=0. k=0.  Mathematics Department, ”Theodor Costescu” National Economic College, Drobeta Turnu - Severin, MEHEDINTI. E-mail address:

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