aa r X i v : . [ m a t h . G M ] F e b On a special kind of improper integral
Alexandru BratosinJanuary 2019
Abstract
Introduction
In the world of mathematical analysis, many counterin-tuitive answers arise from the manipulation of seemingly unrelated con-cepts, ideas, or functions. For example, Euler showed that e iπ + 1 = 0,whereas Gauss proved that the area underneath y = e − x spanning thewhole real axis is √ π . In this paper, we will determine the closed-formsolution of the improper integral I n = Z ∞ ln xx n + 1 dx, ∀ n ∈ R , with n > . Determining closed-form solutions of improper integrals have real implica-tions not only in easing the solving of similar, yet more difficult integrals,but also in speeding up numerical approximations of the answer by makingthem more efficient.
Result
Following our calculations, we derived the formula I n = Z ∞ ln xx n + 1 dx = − π n cot πn csc πn = − ddn " Γ (cid:16) − n (cid:17) Γ (cid:16) n (cid:17) . Depending on the value of n , one may come up with the following intrigu-ing identities: I = Z ∞ ln xx + 1 dx = 0, I = Z ∞ ln xx + 1 dx = − π
27 , I = Z ∞ ln xx + 1 dx = − π √ . . . lim n →∞ I n = − . Introduction
On a special kind of improper integral
To prove the theorem, we will make use of the following definitions and lemmas.Books [1]-[6] should help familiarize with the notions used in this document.
Definition 1 ([3, p. 255]) . The gamma function Γ( z ) is defined as the analyticcontinuation of the factorial to complex arguments, withΓ( z ) = ( z − (Euler integral of the second kind) isΓ( z ) = Z ∞ t z − e − t dt , ∀ z ∈ C , with Re z > , with the subsequent reflection formulaΓ(1 − z )Γ( z ) = π sin πz , ∀ z ∈ C \ Z . Due to the difficulties in analyzing the large and rapidly-increasing functionΓ ′ ( z ), its logarithmic derivative ψ ( z ) is studied instead. Definition 2 ([3, p. 258]) . The digamma function ψ ( z ) is defined as ψ ( z ) = ddz ln Γ( z ) . An integral representation due to Gauss is ψ ( z ) = Z ∞ e − t t e − zt − e − t dt , ∀ z ∈ C , with Re z > . Definition 3 ([3, p. 260]) . The polygamma function ψ ( n ) ( z ) is defined as the n -th derivative of the digamma function, i.e. ψ ( n ) ( z ) = d n dz n ψ ( z ) = d n +1 dz n +1 ln Γ( z ) . Using Gauss’s integral representation of the digamma function ψ ( z ) = ddz ln Γ( z ) = Z ∞ e − t t − e − zt − e − t dt, we differentiate ψ ( z ) n times with respect to z to get the polygamma function ψ ( n ) ( z ) = d n dz n ψ ( z ) = d n dz n Z ∞ e − t t − e − zt − e − t dt , ∀ n ∈ N ∗ . Using Leibniz’s rule for differentiating under the integral sign, we get the integralrepresentation of the polygamma function ψ ( n ) ( z ) = Z ∞ ∂ n ∂z n " e − t t − e − zt − e − t dt = ( − n +1 Z ∞ t n e − zt − e − t dt. Introduction
On a special kind of improper integral
Definition 4.
The trigamma function ψ (1) ( z ) is defined as ψ (1) ( z ) = ddz ψ ( z ) = Z ∞ te − zt − e − t dt. Lemma 1.
For all z ∈ C , with Re z > , Z ∞−∞ t n e − zt − e − t dt = ψ ( n ) (1 − z ) + ( − n +1 ψ ( n ) ( z ) . Proof.
Using the linearity of the integral, we split it into a sum of two integrals. Z ∞−∞ t n e − zt − e − t dt = Z −∞ t n e − zt − e − t dt + Z ∞ t n e − zt − e − t dt. (1)From Definition 3, the right-hand summand of (1) is ( − n +1 ψ ( n ) ( z ) . Z ∞ t n e − zt − e − t dt = ( − n +1 ψ ( n ) ( z ) . (2)For the left-hand summand of (1), substitute u = − t , with du = − dt. Z −∞ t n e − zt − e − t dt = ( − n Z ∞ u n e zu − e u du. Multiply the resulting integrand by e − u e − u . Z −∞ t n e − zt − e − t dt = ( − n Z ∞ u n e − u (1 − z ) − e u du. Factor − Z −∞ t n e − zt − e − t dt = ( − n +1 Z ∞ ue − u (1 − z ) − e − u du. From Definition 3, the right-hand side is ψ ( n ) (1 − z ). Z −∞ t n e − zt − e − t dt = ψ ( n ) (1 − z ) . (3)Therefore, using the results from (2) and (3) in (1), we get Z ∞−∞ t n e − zt − e − t dt = ψ ( n ) (1 − z ) + ( − n +1 ψ ( n ) ( z ) . Introduction
On a special kind of improper integral
Lemma 2.
For all z ∈ C , with Re z > , ψ ( n ) (1 − z ) + ( − n +1 ψ ( n ) ( z ) = ( − n π d n dz n cot πz. Proof.
Using the reflection formula of the gamma functionΓ(1 − z )Γ( z ) = π sin πz , take the natural logarithm of both sides.ln Γ(1 − z ) + ln Γ( z ) = ln π − ln(sin πz ) . Differentiate both sides with respect to z . ddz ln Γ(1 − z ) + ddz ln Γ( z ) = − π cot( πz ) . The above is equivalent to − ψ (1 − z ) + ψ ( z ) = − π cot( πz ) . Differentiate both sides n times with respect to z .( − n +1 ψ ( n ) (1 − z ) + ψ ( n ) ( z ) = − π d n dz n cot( πz ) . ( − n +2 ψ ( n ) (1 − z ) + ( − ψ ( n ) ( z ) = π d n dz n cot( πz ) . ∴ ψ ( n ) (1 − z ) + ( − n +1 ψ ( n ) ( z ) = ( − n π d n dz n cot( πz ) . Lemma 3.
For all x ∈ R , sec x − csc x = − x csc 2 x. Proof.
Using trigonometric identities,sec x − csc x = 1cos x − x = sin x − cos x sin x cos x = − cos 2 x sin x = − x csc 2 x. Main Result
On a special kind of improper integral
Theorem.
For all n ∈ R , with n > , I n = Z ∞ ln xx n + 1 dx = − π n cot πn csc πn = − ddn " Γ (cid:16) − n (cid:17) Γ (cid:16) n (cid:17) . (4) Proof.
To prove this identity, we will calculate I n using the reflection formulaof the polygamma function. First, we’ll introduce the substitution u = n ln x, with du = nx dx. Plugging these into (4) gives us I n = Z ∞ ln xx n + 1 dx = 1 n Z ∞−∞ ue un e u + 1 du. To obtain a convenient form similar to that of the polygamma function, we willdo some basic algebraic manipulation. I n = 1 n Z ∞−∞ ue un e u + 1 du Multiply the integrand by e − u e − u . = 1 n Z ∞−∞ ue u ( n − e − u du Multiply the integrand by 1 − e − u − e − u . = 1 n Z ∞−∞ ue u ( n − (1 − e − u )1 − e − u du. Our original integral is now I n = 1 n Z ∞−∞ ue u ( n − (1 − e − u )1 − e − u du. (5)We will do the following substitution t = 2 u, with dt = 2 du. Main Result
On a special kind of improper integral
Plugging these into (5) yields I n = 1 n Z ∞−∞ ue u ( n − (1 − e − u )1 − e − u du = 14 n Z ∞−∞ te t ( n − (1 − e − t )1 − e − t dt = 14 n " Z ∞−∞ te t ( n − − e − t dt − Z ∞−∞ te t ( n − − e − t dt = 14 n " Z ∞−∞ te t ( n − ) − e − t dt − Z ∞−∞ te t ( n − − e − t dt = 14 n " Z ∞−∞ te − t ( − n ) − e − t dt − Z ∞−∞ te − t (1 − n ) − e − t dt . We end up with the form I n = 14 n " Z ∞−∞ te − t ( − n ) − e − t dt − Z ∞−∞ te − t (1 − n ) − e − t dt . (6)Using Lemma 1, (6) becomes I n = 14 n " ψ (1) (cid:16) − n (cid:17) + ψ (1) (cid:16)
12 + 12 n (cid:17) − ψ (1) (cid:16) − n (cid:17) − ψ (1) (cid:16) n (cid:17) . (7)Using Lemma 2, (7) becomes I n = 14 n " π csc (cid:16) π π n (cid:17) − π csc (cid:16) π n (cid:17) = π n " csc (cid:16) π π n (cid:17) − csc (cid:16) π n (cid:17) = π n " sec (cid:16) π n (cid:17) − csc (cid:16) π n (cid:17) . (8)Using Lemma 3, (8) becomes I n = − π n cot πn csc πn = − ddn " Γ (cid:16) − n (cid:17) Γ (cid:16) n (cid:17) . (cid:4) EFERENCES
On a special kind of improper integral
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