# The Analysis of Contour Integrals

Authors: Tanfer Tanriverdi; JohnBryce Mcleod
Journal: Abstract and Applied Analysis
 (2008-02-03)

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## Abstract

For anyn, the contour integral  y=coshn+1x∮C(cosh(zs)/(sinhz-sinhx)n+1dz,s2=-λ, is associated with differential equation  d2y(x)/dx2+(λ+n(n+1)/cosh2x)y(x)=0. Explicit solutions for n=1 are obtained. For n=1, eigenvalues, eigenfunctions, spectral function, and eigenfunction expansions are explored. This differential equation which does have
solution in terms of the trigonometric functions  does not seem to have been explored and it is also one of the purposes of this
paper to put it on record.


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## Body

## 1. Introduction

When one considers eigenfunction expansions associated with
second-order ordinary differential equations, as Titchmarsh does in his book [1], one is concerned with solutions of the equation
(1.1)−d2y(x)dx2+q(x)y(x)=λy(x)along with
certain boundary conditions, and one tends to say that the only case in which
one can solve this equation explicitly in elementary terms for all λ is the case q(x)=0, when the solutions are of course trigonometric
functions.Now in fact this is not true, and there is in
particular one problem which does not seem to have been explored, and it is the
purpose of this paper to put it on record. Here is the problem:(1.2)d2y(x)dx2+(λ+n(n+1)sech2x)y(x)=0,
which can be
solved explicitly in elementary terms when n is integral.
The explicit solution was known to Kamke [2], but Kamke does not anyway explore the consequences
for eigenfunction expansions nor does Titchmarsh discuss this problem, although
he does discuss problems close to it, for example, (1.3)d2y(x)dx2+(λ+(ν2−14)sec⁡2x)y(x)=0,
on (−π/2,π/2), which leads, when ν=n, to an expansion in series involving associated Legendre
functions.It is perhaps worth remarking how our interest in this
problem arises. In [3] there is the question of
travelling waves and steady solutions for a discrete reaction-diffusion equation
of the type
(1.4)un′=un+1−2un+un−1+f(un),
where the
function f is “bistable”. That is, there exist three
numbers U1,  U2,  U3,  U1<U2<U3, such that
(1.5)f(U1)=f(U2)=f(U3)=0 with f<0 in (U1,U2) and f>0 in (U2,U3). A prototypical f would be
(1.6)f(u)=−Asin⁡(u)+F,
where A and F
are positive constants and F<A, so that
(1.7)U1=sin⁡−1(FA),U2=π+sin⁡−1(FA),U3=2π+sin⁡−1(FA). Such equations
arise in a number of different applications, for example, in dislocation theory
where un is the
displacement of the nth atom in some
material, or in neurobiology where un is typically
the electric potential of the nth nerve cell, and
in both these applications the interest is in monotonic solutions un with un→U1 as n→−∞, un→U3 as n→∞ [3–6].The basic question is whether there exist such
solutions with the form of a travelling wave,u(n−ct), c≠0, or of a steady solution or standing wave, where c=0, and there is an important distinction between these
two cases. For a travelling wave, c≠0, un is clearly a
function of the continuous variable t, and indeed because of (1.4), a differentiable function of t. This leads to the difference-differential equation
(1.8)ut(x,t)=u(x+1,t)−2u(x,t)+u(x−1,t)+f(u(x,t)). If, however, c=0, then, as in [3], we have to study the purely difference equation
(1.9)u(x+1,t)−2u(x,t)+u(x−1,t)+f(u(x,t))=0,
and the
solutions may be discontinuous since there is nothing that now connects values
of u(x) with values of u(x+δ) for |δ|<1. It is best therefore to think of the solution of (1.9) as a number of
(monotonic) sequences unα indexed by α, each satisfying
(1.10)un+1−2un+un−1+f(un)=0. The simplest
case would be that there is just one such sequence (modulo the translation n→n+k, k integral), but
it is possible that there may be a finite number, or even a partial or total
continuum.In view of applications, where the distance between
atoms or nerve cells is small, it is more natural to think of (1.8) in the form
(1.11)ut(x,t)=u(x+ϵ,t)−2u(x,t)+u(x−ϵ,t)+f(u(x,t)),
where ϵ is small and
represents the distance between atoms or nerve cells. A tempting approximation
is then
(1.12)ut(x,t)∼ϵ2uxx(x,t)+f(u(x,t)),
and in order to
make sense of the scaling, in [3] the authors
introduced a factor ϵ2 in front of f. This therefore leads to a comparison between the
solutions of
(1.13)un′=un+1−2un+un−1+ϵ2f(un)=0,(1.14)ut(x,t)=uxx(x,t)+f(u). For the
continuous diffusion problem, the answer is both simple and well known [4, 5].Given a functionf that is
bistable, there is just one possible wave-speed c, and this value of c is 0, that is, there is a steady solution if and only if
(1.15)∫U1U3f(u)ⅆu=0. (The proof is a
simple phase plane argument, and  c=0 implies (1.15) follows by
multiplying (1.14) by u′ and
integrating.)The solution in the discrete case is however
different, as  discussed in [6]. There may continue to be steady solutions where (1.15) no longer
holds. Consider specifically the case (1.6), so that
(1.16)un′=un+1−2un+un−1−ϵ2sin⁡un+F. The case corresponding to (1.15) is F=0,  but the authors, in [3], have shown that for F sufficiently
small, say |F|<Fcrit, there exist precisely two steady
solutions of (1.16), and Fcrit, which of course depends on ϵ, can be evaluated for small ϵ. Specifically,
(1.17)Fcrit∼Be−π2/ϵ,
where the
constant B=64π∫0π(sin⁡2(s)/s)ds is given. For |F|>Fcrit, the solutions move and the equation has travelling
wave solutions instead of steady solutions.In order to prove results such as (1.17), one has to regard (1.16) as a singular
perturbation of the steady continuous-diffusion equation
(1.18)uxx−sin⁡u=0,
for which the
solution (satisfying u(−∞)=0, u(∞)=2π) is U=4tan⁡−1ex. If we linearize (1.18) about U, we obtain
(1.19)ϕ′′−cos⁡(U)ϕ=0. But multiplying (1.18) by U′ and integrating
lead to (1/2)(U′)2=1−cos⁡U, so that since U′=2sech(x), we have cosU=1−2sech2(x). The linearization (1.19) thus becomes ϕ′′+(−1+2sech2x)ϕ=0, which is of course (1.2) with λ=−1 and n=1. Thus the selfadjoint operator T given (in L2(−∞,∞)) by Tϕ=−ϕ′′−2sech2(x)ϕ has an
eigenvalue at −1, with eigenfunction U′ (differentiation of (1.18) shows that U′ satisfies (1.19)). This fact,
together with the additional fact that the spectrum of T is continuous
above 0 (since sech2(x)∈L(0,∞)), is highly
relevant to the work in [3] and led to our
interest more generally in the spectral problem (1.2).The explicit solution for anyn using contour
integrals different from what Kamke did is known to [7]. For more information on this problem one can see [7, 8].


## 2. Preliminaries

We want to know
expansion of an arbitrary functionf(x) in terms of
eigenfunctions. So one needs to know the following. Let θ(x,λ) and ϕ(x,λ) be the
solutions of (1.1) such that
(2.1)ϕ(0,λ)=sin⁡α,ϕ′(0,λ)=−cos⁡α,θ(0,λ)=cos⁡α,θ′(0,λ)=sin⁡α,
where α is real. Wx(ϕ,θ)=W0(ϕ,θ)=1. The general solution of (1.1) is of the form
(2.2)ψ(x,λ)=θ(x,λ)+m(λ)ϕ(x,λ)∈L2(0,∞). The spectrum is
defined by means of the function
(2.3)k(λ)=lim⁡δ→0∫0λ−ℑ{m(u+iδ)}du,
which exists
for all real λ and k(λ) is a
nondecreasing function. The expansion of a function f(x) in terms of the
spectral function depends on the following lemmas taken from [1].Lemma 2.1.
Without detailing, let the interval be(0,∞):
(2.4)f(x)=1π∫0∞ϕ(x,λ)dk(λ)∫0∞ϕ(t,λ)f(t)dt. If m(λ) has poles, then
(2.5)f(x)=∑N=0∞ϕN(t)∫0∞ϕN(t,λ)f(t)dt.Lemma 2.2.
Without detailing, let the interval be(−∞,∞). If q(x)
is an even
function, then m1(λ)=−m2(λ). So the expansion formula is
(2.6)f(x)=1π∫−∞∞θ(x,λ)dξ∫−∞∞θ(y,λ)f(y)dy+1π∫−∞∞ϕ(x,λ)dζ∫−∞∞ϕ(y,λ)f(y)dy,
where
(2.7)ξ(λ)=lim⁡δ→0∫0λ−ℑ{1m1(u+iδ)−m2(u+iδ)}du,ξ′(λ)=ℑ{12m2(λ)},ζ(λ)=lim⁡δ→0∫0λ−ℑ{m1(u+iδ)m2(u+iδ)m1(u+iδ)−m2(u+iδ)}du,ζ′(λ)=−12ℑ{m2(λ)}.


## 3. Main Results

We are now
dealing with (1.2) in the case
where n is integral.
Without loss of generality, we may suppose n≥0, but since n=0 reduces (1.2) to the simple
trigonometric case, we are in fact interested only in n>0. We first prove that a solution is given by
(3.1)y=cosh⁡n+1x∮Ccosh⁡(zs)(sinh⁡z−sinh⁡x)n+1dz,s2=−λ,
where the
contour C is taken round
the point z=x and no other
zero of sinh⁡z−sinh⁡x. This is slight variant of a form which Titchmarsh uses
in his discussion of (1.3). The proof below will show  (3.1), being continuous at
least formally, to be a solution of (1.2) where n is not an
integer, but the difficulty then is to choose a suitable contour, since the
integrand has a branch point at z=x.Remark 3.1.
We also
remark that it is obvious that we can express the solution (3.1)
equivalently
ignoring some multiplicative constants as
(3.2)y(x)=cosh⁡n+1xdn(cosh⁡xdx)n∮Ccosh⁡(zs)sinh⁡z−sinh⁡xdz.Theorem 3.2.
The
contour integral (3.1)
satisfies the
differential equation (1.2).Proof.
We
see that(3.3)y′=(n+1)tanh⁡(x)y+(n+1)cosh⁡n+2x∮Ccosh⁡(zs)(sinh⁡z−sinh⁡x)n+2dz,(3.4)y′′+n(n+1)sech2(x)y=(n+1)cosh⁡n+1x∮Ccosh⁡(zs){(n+1)sinh⁡2z+sinh⁡zsinh⁡x+n+2}(sinh⁡z−sinh⁡x)n+3dz. Integrating (3.1) by parts,
(3.5)y(x)=n+1s2∮Ccosh⁡(zs){(n+1)sinh⁡2z+sinh⁡zsinh⁡x+n+2}(sinh⁡z−sinh⁡x)n+3dz,
so that
(3.6)λy=−(n+1)cosh⁡n+1x∮Ccosh⁡(zs){(n+1)sinh⁡2z+sinh⁡zsinh⁡x+n+2}(sinh⁡z−sinh⁡x)n+3dz. Comparing (3.4) and (3.6), we see that y′′+n(n+1)sech2(x)y=−λy, so that y(x) satisfies (1.2), as required.Remark 3.3.
We now point
out that the factorcosh⁡(zs)
played little
part in the argument. Certainly, the argument would have washed equally well if
we had replaced cosh⁡(zs) by sinh⁡(zs):
(3.7)y2(x)=cosh⁡n+1x∮Csinh⁡(zs)(sinh⁡z−sinh⁡x)n+1dz.Theorem 3.4.
The
contour integral (3.7)
satisfies the
differential equation (1.2).Proof.
Proof is the same as the above theorem. So we omit it.Remark 3.5.
Furthermore,
once the integrands have poles atz=x, the solution can be evaluated by calculating the
relevant residues. For example, in the trivial case n=0, when we should recover the trigonometric functions,
the residues of
(3.8)cosh⁡(zs)sinh⁡z−sinh⁡x
are (3.9)cosh⁡(xs)cosh⁡x,
so that the
solution (3.7) becomes
multiples of cos⁡(xλ) (similarly sin⁡(xλ)), as we
expect.
We can
generalize Theorems (3.2) and (3.4) by defining the
following operator:(3.10)Tf(x)=cosh⁡n+1x∮Cf(z)(sinh⁡z−sinh⁡x)n+1dz,where f is a
differentiable function as long as one can pick up residue.Corollary 3.6.
Iff(z)=cosh⁡(zs)(sinh⁡(zs)), then we obtain
Theorems (3.2) and (3.4). The operator T
is also  linear.


## 4. The Explicit Solution Given by Residues forn=1

We now require
the residues of(4.1)cosh⁡(zs)(sinh⁡z−sinh⁡x)2,sinh⁡(zs)(sinh⁡z−sinh⁡x)2.  Since (4.2)cosh⁡(zs)(sinh⁡z−sinh⁡x)2={cosh⁡(xs)+(z−x)ssinh⁡(xs)+⋯}{1−(z−x)tanh⁡(x)+⋯}(z−x)2cosh⁡2(x), we see that the
residue at z=x is(4.3)ssinh⁡(xs)−tanh⁡(x)cosh⁡(xs)cosh⁡2(x),  so that one
solution is(4.4)y1=λsin⁡(xλ)+tanh⁡(x)cos⁡(xλ).  By examining
the residue of the second equation of (4.1), we see that a second solution is(4.5)y2=λcos⁡(xλ)−tanh⁡(x)sin⁡(xλ).Remark 4.1.
The solution
can also be obtained from (3.2). For we have already seen, from our brief discussion
of the case n=0, that the integral in (3.2)
is just a
multiple of cos⁡(xλ)/cosh⁡x (or of sin⁡(xλ)/cosh⁡x if we replace cosh⁡(zs) by sinh⁡(zs))    hence in the
first case, (3.2)
gives a
multiple of
(4.6)cosh⁡(x)ddx(cos⁡(xλ)cosh⁡(x))=−λsin⁡(xλ)−tanh⁡(x)cos⁡(xλ),in accordance
with (4.4).Remark 4.2.
WronskianW(y1(x),y2(x))=−λ(λ+1). We now have two linearly independent solutions.Remark 4.3.
The general solution isy(x)=c1y1(x)+c2y2(x).Lemma 4.4.
y1(x)
is an odd
function but y2(x)
is an even
function.Proof.
(4.7)y1(−x)=−y1(x),y2(−x)=y2(x).


## 5. Eigenvalues and Eigenfunctions forn=1 When y(0)=y(b)=0

Theorem 5.1.
Eigenvalues with associated boundary conditionsy(0)=y(b)=0
are the zeros
of
(5.1)λtan⁡(bλ)+tanh⁡(b)=0;
furthermore,
one and only one eigenvalue lies in the interval
(5.2)(k−12)π<bλ<(k+12)π
for every
integral k≠0.Proof.
One
can see from Remark4.3 that if y(0)=0, then c2=0. So that y(b)=0 implies
(5.3)λsin⁡(bλ)+tanh⁡(b)cos⁡(bλ)=0,λtan⁡(bλ)=−tanh⁡(b). One can see
immediately that the eigenvalues belong to the interval (5.2).
To prove the second part we use the following
strategy. Multiply (5.1) by b and set x=bλ, then denote h(x)=xtan⁡(x)+btanh⁡(b). So h′(x)=sec⁡2(x){(1/2)sin⁡(2x)+x}. Notice that h(x) is an even and
does not intersect x-axis, where −1/2<x<1/2. If x<−1/2, then h′(x)<0 and if x>1/2, then h′(x)>0. So the monotonicity of h(x) implies that h(x) has only one
zero belonging to the interval (5.2) for every
integral k≠0.Remark 5.2.
Soyk is the
eigenfunction in the form of (4.4):
(5.4)yk=λksin⁡(xλk)+tanh⁡(x)cos⁡(xλk).Corollary 5.3.
One can
orthonormalize the eigenfunctions.Proof.
(5.5)∫0byk2dx=−4λktanh⁡(b)cos⁡2(bλk)+2λk(λk+1)b−(λk−1)sin⁡(2bλk)4λk. The
orthonormalized eigenfunctions denoted by Φ(x,λk), Φ(x,λk)=(λksin⁡(xλk)+tanh⁡(x)cos⁡(xλk))/∫0byk2dx.Remark 5.4.
An arbitrary
functionf(x)
in terms of
eigenfunctions follows:
(5.6)f(x)=∑k=0∞CkΦ(x,λk),⁢whereCk=∫0bf(x)Φ(x,λk)dx.


## 6. Eigenvalues and Eigenfunctions forn=1 When y′(0)=y′(b)=0

Theorem 6.1.
The
eigenvalues withy′(0)=y′(b)=0
are the zeros
of (6.1); furthermore, there exists one and only one
eigenvalue lying in the interval (5.2)
for every
integral k≠0.Proof.
One
can see from Remark4.3 that if y′(0)=0, then either c1=0 or λ=−1. If λ=−1, then the
associated eigenfunction is zero. So this is useless. Hence, c1=0. y′(b)=0 implies
(6.1)λtan⁡(bλ)+sech2(b)tan⁡(bλ)+λtanh⁡(b)=0.So it is
obvious that the zeros (eigenvalues) belong to the interval (5.2).
Setx=bλ. (b2λ=x2λ=x2/b2). Equation (6.1) is denoted by h(x):
(6.2)h(x)=(x2b2+sech2(b))tan⁡(x)+xbtanh⁡(b). It is enough to
show that h(x) is monotonic:
(6.3)h′(x)=x2+b2sech2(b)+xsin⁡(2x)b2cos⁡2(x)+1btanh⁡(b). We see that h′(x)>0 everywhere. We
therefore conclude that h(x) is monotonic.Remark 6.2.
The
associated eigenfunctions are(6.4)yk=λkcos⁡(xλk)−tanh⁡(x)sin⁡(xλk).Corollary 6.3.
One can orthonormalize the eigenfunctions.Proof.
(6.5)∫0byk2dx=−4λktanh⁡(b)sin⁡2(bλk)+2λk(λk+1)b+(λk−1)sin⁡(2bλk)4λk. The
orthonormalized eigenfunctions denoted by Ψ(x,λk), Ψ(x,λk)=(λkcos⁡(xλk)−tanh⁡(x)sin⁡(xλk))/∫0byk2dx.Remark 6.4.
Therefore, an arbitraryf(x)
in terms of
orthonormalized eigenfunctions is
(6.6)f(x)=∑k=0∞CkΨ(x,λk),Ck=∫0bf(x)Ψ(x,λk)dx.


## 7. Spectral Functionm(λ) over (0,∞) and Expansion

Now letθ(x,λ) and ϕ(x,λ) be the
solutions of
(7.1)d2ydx2+{λ+2sech2(x)}y=0,
which satisfy (2.1); so that
(7.2)ϕ(x,λ)=−cos⁡α{λsin⁡(xλ)+tanh⁢⁡xcos⁡(xλ)}λ+1+sin⁡α{λcos⁡(xλ)−tanh⁢⁡xsin⁡(xλ)}λ,(7.3)θ(x,λ)=sin⁡α{λsin⁡(xλ)+tanh⁢⁡xcos⁡(xλ)}λ+1+cos⁡α{λcos⁡(xλ)−tanh⁢⁡xsin⁡(xλ)}λ. Now we need to
find m(λ). This suggests that
(7.4)ψ(x,λ)=θ(x,λ)+m(λ)ϕ(x,λ)∈L2(0,∞). To get demanded
result, one needs to find the asymptotics of θ(x,λ) and ϕ(x,λ) as x→∞ and ℑλ>0:
(7.5)ϕ(x,λ)∼{−λcos⁡α+λ(λ+1)sin⁡α+i(−λcos⁡α−(λ+1)sin⁡α)}e−ixλ2λ(λ+1)=M1(λ)e−ixλ,θ(x,λ)∼{λsin⁡α+λ(λ+1)cos⁡α+i(λsin⁡α−(λ+1)cos⁡α)}e−ixλ2λ(λ+1)=M(λ)e−ixλ.
Finally, we must arrange the linear combination so that
the terms e−ixλ cancel. That
is, M(λ)+m(λ)M1(λ)=0. Hence,
(7.6)m(λ)=−iλ(λ+1)−(λ2+λ+1)sin⁡αcos⁡αλcos⁡2α+(λ+1)2sin⁡2α,ℑm(λ)={−λ⁢⁢(λ+1)λcos⁡2α+(λ+1)2sin⁡2αwhenλ>0,0whenλ<0. We see that the
spectrum is continuous for λ>0. But we have a point spectrum for λ<0. So that an arbitrary f(x) is a linear
combination of integrand and series. The spectral function is calculated from (2.3). Hence, (7.7)dk(λ)={λ(λ+1)λcos⁡2α+(λ+1)2sin⁡2αwhen⁢λ>0,0when⁢λ<0. In particular,
if α=0, then(7.8)dk(λ)={λ+1λwhen⁢λ>0,0when⁢λ<0. So that in this
case there is no eigenvalue for λ<0. From Lemma 2.1, one can see the
expansion of(7.9)f(x)=1π∫0∞λsin⁡(xλ)+tanh⁢⁡xcos⁡(xλ)λdλ∫0∞λsin⁡(yλ)+tanh⁢⁡ycos⁡(yλ)λ+1f(y)dy. Similarly, if α=π/2, then (7.10)ℑm(λ)={−λλ+1when⁢λ>0,0when⁢λ<0.. Hence, there
exits only one eigenvalue at λ=−1 and the
corresponding eigenfunction ϕ(x,−1) where ϕ(x,λ) is (7.2). So the spectrum calculated by (2.3) is(7.11)dk(λ)={λλ+1when⁢λ>0,0when⁢λ<0..   Therefore, from
Lemma 2.1, an expansion of function f(x) in terms of
eigenfunctions and spectral function follows:(7.12)f(x)=1π∫0∞λcos⁡(xλ)−tanh⁢⁡xsin⁡(xλ)λ+1dλ×∫0∞λcos⁡(yλ)−tanh⁢⁡ysin⁡(yλ)λf(y)dy+c1sech(x), where c1 is a constant.
So we have proved the following theorem regarding the nature of m(λ).Theorem 7.1.
Ifα=0, there is no
eigenvalue where λ<0. If α=π/2, then there exists only one eigenvalue at λ=−1 and the
corresponding eigenfunction is ϕ(x,−1)=sech(x).Finally, one
can ask what is the range ofα in the case of λ<0 and n=1.Theorem 7.2.
If0<α<π/2, then there are precisely two eigenvalues except at α=π/4. Hence there are two eigenfunctions, namely, ϕ(x,λ1) and ϕ(x,λ2).Proof.
We
check the zeros of both the numerator and denominator of (7.6). After working out the algebra, we see that the zeros
of the numerator and the denominator are (7.13)μ1=−(2+tan⁡2α)+tan⁡α4+tan⁡2α+12,μ2=−(2+tan⁡2α)−tan⁡α4+tan⁡2α+12,λ1=−(2tan⁡2α+1)+4tan⁡2α+12tan⁡2α,λ2=−(2tan⁡2α+1)−4tan⁡2α+12tan⁡2α.
If m(λ) has poles, then
they are the eigenvalues. If so, the eigenfunctions are ϕ(x,λ1) and ϕ(x,λ2). Finally, there is one thing to be proved in this
case α=π/4. To do this, we expand the zeros around α=π/4 and at the end α→π/4: (7.14)λ1=−3+52+20−12510(α−π4)+−400+3525100(α−π4)2+O{(α−π4)3},λ2=−3−52+20+12510(α−π4)+−400−3525100(α−π4)2+O{(α−π4)3}.
Similarly, (7.15)μ1=−3+62+76−126(α−π4)+−288+118636(α−π4)2+O{(α−π4)3},μ2=−3−62+−76−126(α−π4)+−288−118636(α−π4)2+O{(α−π4)3}.  That completes
the proof.


## 8. Expansion over(−∞,∞)

Now consider
the interval(−∞,∞) instead of (0,∞). From  (2.7)  the following
can be calculated:
(8.1)ξ′(λ)={λ(λ+1)2λsin⁡2α+2(λ+1)2cos⁡2αwhen⁢λ>0,0when⁢λ<0,ζ′(λ)={(λ+1)λ2λcos⁡2α+2(λ+1)2sin⁡2αwhen⁢λ>0,0when⁢λ<0.
Now one can use (7.2), (7.3), and (8.1) gets the
following expansion from Lemma (2.2):
(8.2)f(x)=1π{∫0∞λcos⁡(xλ)−tanh⁢⁡xsin⁡(xλ)2(λ+1)dλ∫−∞∞λcos⁡(yλ)−tanh⁢⁡ysin⁡(yλ)λf(y)dy+∫0∞λsin⁡(xλ)+tanh⁢⁡xcos⁡(xλ)2λdλ∫−∞∞λsin⁡(yλ)+tanh⁢⁡ycos⁡(yλ)λ+1f(y)dy}+c1sech(x), where c1 is a constant.



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*Source: 765920-2008-03-12.xml*