Integrals and Miscellaneous 3

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Integrals and Miscellaneous 1

Integrals and Miscellaneous 2

2022/4/7

I=0(1+x4x2)sdxI=0(1+x4x2)sdxLet y=(1+x4x2)2y=(1+x4x2)2Then,

I=14210[ys254(1y)12+2ys214(1y)12]dy=142[B(s214,32)+2B(s2+34,12)]=142[Γ(s214)π2Γ(s2+54)+2Γ(s2+34)πΓ(s2+54)]=π2(12+s12)Γ(s214)(2s+1)Γ(s2+14)=s2s+1π2Γ(s214)Γ(s2+14)

2022/4/7

I=0sinsxe2πx1dxAbel-Plana formula:nk=mf(k)=f(m)+f(n)2+nmf(z)dz+i0f(m+iy)f(miy)f(n+iy)+f(niy)e2πy1dy

Let f(z)=esz , m=0 , n= , we havek=0esk=12+0esxdx+i0eisyeisye2πy1dyTherefore,11es=12+1s+20sinsxe2πx1dxHence,I=12(11es1s12)


Another solution:

I=0sinsxn=1e2πnxdx=n=10e2πnxsinsxdx

This integral can be evaluated like this:0e2πnxsinsxdx=[e2πnx2πnsinsx]0+s2πn0e2πnxcossxdx=[se2πnx4π2n2cossx]0s24π2n20e2πnxsinsxdxHence,0e2πnxsinsxdx=ss2+4π2n2

I=n=1ss2+4π2n2=14(coths22s)=12(11es1s12)

2022/4/9 A

I=0xs1eixdxConsider 0=Czs1eizdzContour 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=0xs1eixdx+π20iRseisθeiRcosθRsinθdθeπ2si0ys1eydyThe second integral tends to 0 as R when Rs<1.

So we suppose that Rs<1.Then,I=eπ2si0ys1eydy=Γ(s)eπ2siHence,{0xs1cosxdx=Γ(s)cosπs20xs1sinxdx=Γ(s)sinπs2

2022/4/9 B

From @integralsbotπ20dx31+3sin2x=Γ3(13)273πPROOF.LHS=π20dx31+3sin2x=π20n=0(13)n(3sin2x)nn!dx=n=0(13)nn!(3)nπ20sin2nxdx=π2n=0(13)n(12)n(1)n(3)nn!=π2F(13,12,1;3)Recall Bailey's identityF(a,b,2b;4z(1z)2)=(1z)2aF(a,ab+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,1a2,3a+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,π20dx31+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.

2022/4/9 C

I=0cos2bxcoshaxdx=20eaxcos2bx1+e2axdx=20eaxcos2bxn=0(e2ax)ndx=2n=0(1)n0e(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=2an=0(1)n(2n+1)(2n+1)2a2+4b2=π22an=0(1)n(2n+1)π2(n+12)2+b2a2π2Using the series representation of sechz:sechz=πn=0(1)n(2n+1)π2(n+12)2+z2We obtainI=π2asechbaπ

2022/4/10

Iπ20arctantanxtanxdx

proof here:

2022/4/10

I=1+x21+x4dx=12(1x2+2x+1+1x22x+1)dx=12dx(x+12)2+12+12dx(x12)2+12=arctan(2x+1)+arctan(2x1)2+C=12arctan2x1x2+C

2022/4/10

I=0arctanxxlogx+1dx=10arctanxxlogx+1dx+1arctanxxlogx+1dxReplace x1x in the second term ,then since arctan1x=π2arctanx ,I=π210dxxlogx+1Substitute x=ey, we haveI=π20ey2dyTherefore,0arctanxxlogx+1dx=π324

2022/4/11

I=0(121x+1ex1)exxdxBinet's first formula:logΓ(z)=(z12)logzz+12log2π+0(1et11t+12)eztdttSubstitute z=1, we obtain0=1+12log2π+IHence,I=112log2π


Another solution:I=0(121x+1ex1)exxdx

Define μ(x) in x>0 by μ(x)logΓ(x)(x12)logx+x12log2πWe can see below easilyμ(x)μ(x+1)=(x+12)log(11x+1)+1Using log(1X)=XX22X33, we haveμ(x)μ(x+1)=n=12x+12n(x+1)n1=n=1(1n(x+1)n112n(x+1)n)=12n=1n(n+1)(n+2)(x+1)n+1Take the sum of x , x+1 , x+2 , ,x+N1 and use limNμ(x+N)=0, we getμ(x)=12n=1[n(n+1)(n+2)k=01(x+k+1)n+1]=12n=1[n(n+1)(n+2)1n!k=00tne(x+k+1)tdt]=12n=1[(2n+21n+1)1n!0exttnet1dt]=120[2t2n=0tn+2n!(n+2)1tn=0tn+1(n+1)!]extdtet1=0(121t+1et1)exttdtHence,I=μ(1)=112log2π

2022/4/19

I=0xdx3(e3x1)2=1910t13(1t)23logtdt(e3x=t)=19B(x,y)x|x=23,y=13=19B(23,13)[ψ(23)ψ(1)]=19πsinπ3(γ32log3+π23+γ)=π33(log3π33)

2022/4/19
2022/4/20

Thanks to @BenriBot_ccbs. He told me about an amazing substitution! I10arctanxx(1x2)dxSubstituting x1x1+x, we haveI=10arctan1x1+xx(1x2)dxAdding these two Is,2I=10arctanx+arctan1x1+xx(1x2)dx=10π4x(1x2)dxHence,I=π810dxx(1x2)=π1610y34(1y)12dy(x=y)=π16B(14,12)=π16πΓ(14)Γ(34)=2πΓ2(14)32

If we expand to the series,I=1210y34(1y)12arctanydy=1210y34(1y)12n=0(1)n2n+1yn+12dy=12n=0(1)n2n+110yn14(1y)12dy=12n=0(1)n2n+1B(n+34,12)=π2n=0(1)n2n+1Γ(n+34)Γ(n+54)=2πΓ(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)4π

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Integrals and Miscellaneous 4

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