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I=∫∞0(√1+x4−x2)sdxI=∫∞0(√1+x4−x2)sdxLet y=(√1+x4−x2)2y=(√1+x4−x2)2Then,
I=14√2∫10[ys2−54(1−y)12+2ys2−14(1−y)−12]dy=14√2[B(s2−14,32)+2B(s2+34,12)]=14√2[Γ(s2−14)√π2Γ(s2+54)+2Γ(s2+34)√πΓ(s2+54)]=√π2(12+s−12)Γ(s2−14)(2s+1)Γ(s2+14)=s2s+1√π2Γ(s2−14)Γ(s2+14)
I=∫∞0sinsxe2πx−1dxAbel-Plana formula:n∑k=mf(k)=f(m)+f(n)2+∫nmf(z)dz+i∫∞0f(m+iy)−f(m−iy)−f(n+iy)+f(n−iy)e2πy−1dy
Let f(z)=e−sz , m=0 , n=∞ , we have∞∑k=0e−sk=12+∫∞0e−sxdx+i∫∞0e−isy−eisye2πy−1dyTherefore,11−e−s=12+1s+2∫∞0sinsxe2πx−1dxHence,I=12(11−e−s−1s−12)
Another solution:
I=∫∞0sinsx∞∑n=1e−2πnxdx=∞∑n=1∫∞0e−2πnxsinsxdx
This integral can be evaluated like this:∫∞0e−2πnxsinsxdx=[−e−2πnx2πnsinsx]∞0+s2πn∫∞0e−2πnxcossxdx=[−se−2πnx4π2n2cossx]∞0−s24π2n2∫∞0e−2πnxsinsxdxHence,∫∞0e−2πnxsinsxdx=ss2+4π2n2
I=∞∑n=1ss2+4π2n2=14(coths2−2s)=12(11−e−s−1s−12)
See below about Abel-Plana summatioin formula:
See below about a series formula involving cothx:
I=∫∞0xs−1eixdxConsider 0=∮Czs−1eizdzContour C starts at the origin, go right along the real axis to R, then draw the quadrant of radius R counter-clockwise, go back to the origin along the imaginary axis.0=∫∞0xs−1eixdx+∫π20iRseisθeiRcosθ−Rsinθdθ−eπ2si∫∞0ys−1e−ydyThe second integral tends to 0 as R→∞ when Rs<1.
So we suppose that Rs<1.Then,I=eπ2si∫∞0ys−1e−ydy=Γ(s)eπ2siHence,{∫∞0xs−1cosxdx=Γ(s)cosπs2∫∞0xs−1sinxdx=Γ(s)sinπs2
From @integralsbot∫π20dx3√1+3sin2x=Γ3(13)273πPROOF.LHS=∫π20dx3√1+3sin2x=∫π20∞∑n=0(13)n(−3sin2x)nn!dx=∞∑n=0(13)nn!(−3)n∫π20sin2nxdx=π2∞∑n=0(13)n(12)n(1)n(−3)nn!=π2F(13,12,1;−3)Recall Bailey's identityF(a,b,2b;−4z(1−z)2)=(1−z)2aF(a,a−b+12,b+12;z2)Let z=13 , a=13 , b=12 to getF(13,12,1;−3)=(23)23F(13,13,1;19)Moreover, recall a formula deduced here, namelyF(a,1−a2,3a+56;19)=(34)a2Γ(3a+56)Γ(23)Γ(3a+46)Γ(56)After setting a=13, duplication formula and reflection formula givesF(13,13,1;19)=323Γ3(13)4π2 Therefore,∫π20dx3√1+3sin2x=Γ3(13)273π
We can find (1) in:
Bailey.W.N, Products of generalized hypergeometric series (1928)
Brychkov.Y.A, Handbook of Special Functions: Derivatives, Integrals, Series and Other Formulas (2008)
Bateman.H, Eldélyi.A, Higher Transcendental Functions (1953)
And (2) in:
Brychkov.Y.A, Handbook of Special Functions: Derivatives, Integrals, Series and Other Formulas (2008)
Karlsson.P.W, On two hypergeometric summation formulas conjectured by Gosper (1986)
@AlbahariRicardo obtained the same result in so ellegant way. See here.
I=∫∞0cos2bxcoshaxdx=2∫∞0e−axcos2bx1+e−2axdx=2∫∞0e−axcos2bx∞∑n=0(−e−2ax)ndx=2∞∑n=0(−1)n∫∞0e−(2n+1)axcos2bxdxThe rightmost side is evaluated by integration by parts twice then,∫∞0e−(2n+1)axcos2bxdx=(2n+1)a(2n+1)2a2+4b2Hence,I=2a∞∑n=0(−1)n(2n+1)(2n+1)2a2+4b2=π22a∞∑n=0(−1)n(2n+1)π2(n+12)2+b2a2π2Using the series representation of sechz:sechz=π∞∑n=0(−1)n(2n+1)π2(n+12)2+z2We obtainI=π2asechbaπ
I=∫1+x21+x4dx=12∫(1x2+√2x+1+1x2−√2x+1)dx=12∫dx(x+1√2)2+12+12∫dx(x−1√2)2+12=arctan(√2x+1)+arctan(√2x−1)√2+C=1√2arctan√2x1−x2+C
I=∫∞0arctanxxlogx+1dx=∫10arctanxxlogx+1dx+∫∞1arctanxxlogx+1dxReplace x→1x in the second term ,then since arctan1x=π2−arctanx ,I=π2∫10dxxlogx+1Substitute x=e−y, we haveI=π2∫∞0e−y2dyTherefore,∫∞0arctanxxlogx+1dx=π324
I=∫∞0(12−1x+1ex−1)e−xxdxBinet's first formula:logΓ(z)=(z−12)logz−z+12log2π+∫∞0(1et−1−1t+12)e−ztdttSubstitute z=1, we obtain0=−1+12log2π+IHence,I=1−12log2π
Proof of Binet's first formula:
Another solution:I=∫∞0(12−1x+1ex−1)e−xxdx
Define μ(x) in x>0 by μ(x)≡logΓ(x)−(x−12)logx+x−12log2πWe can see below easilyμ(x)−μ(x+1)=(x+12)log(1−1x+1)+1Using log(1−X)=−X−X22−X33−⋯, we haveμ(x)−μ(x+1)=∞∑n=12x+12n(x+1)n−1=∞∑n=1(1n(x+1)n−1−12n(x+1)n)=12∞∑n=1n(n+1)(n+2)(x+1)n+1Take the sum of x , x+1 , x+2 , ⋯,x+N−1 and use limN→∞μ(x+N)=0, we getμ(x)=12∞∑n=1[n(n+1)(n+2)∞∑k=01(x+k+1)n+1]=12∞∑n=1[n(n+1)(n+2)1n!∞∑k=0∫∞0tne−(x+k+1)tdt]=12∞∑n=1[(2n+2−1n+1)1n!∫∞0e−xttnet−1dt]=12∫∞0[2t2∞∑n=0tn+2n!(n+2)−1t∞∑n=0tn+1(n+1)!]e−xtdtet−1=∫∞0(12−1t+1et−1)e−xttdtHence,I=μ(1)=1−12log2π
detailed text here:
I=∫∞0xdx3√(e3x−1)2=−19∫10t−13(1−t)−23logtdt(e−3x=t)=−19∂B(x,y)∂x|x=23,y=13=−19B(23,13)[ψ(23)−ψ(1)]=−19πsinπ3(−γ−32log3+π2√3+γ)=π3√3(log3−π3√3)
About ψ(2/3):
About the derivative of the beta function:
I=∫∞0x(ex−1)23dx
Thanks to @BenriBot_ccbs. He told me about an amazing substitution! I≡∫10arctanx√x(1−x2)dxSubstituting x→1−x1+x, we haveI=∫10arctan1−x1+x√x(1−x2)dxAdding these two Is,2I=∫10arctanx+arctan1−x1+x√x(1−x2)dx=∫10π4√x(1−x2)dxHence,I=π8∫10dx√x(1−x2)=π16∫10y−34(1−y)−12dy(x=√y)=π16B(14,12)=π16√πΓ(14)Γ(34)=√2πΓ2(14)32
If we expand to the series,I=12∫10y−34(1−y)−12arctan√ydy=12∫10y−34(1−y)−12∞∑n=0(−1)n2n+1yn+12dy=12∞∑n=0(−1)n2n+1∫10yn−14(1−y)−12dy=12∞∑n=0(−1)n2n+1B(n+34,12)=√π2∞∑n=0(−1)n2n+1Γ(n+34)Γ(n+54)=2√πΓ(34)Γ(14)3F2[1,12,3432,54;−1]=π43F2[12,12,3432,1;1]Then we get an identity about 3F2,3F2[12,12,3432,1;1]=√2Γ2(14)4√π
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