Saturday, 2 May 2015

Definite integrals 

∫∞0dxx2+a2=π2a
∫∞0xp−1dx1+x=πsin(pπ), 0<p<1
∫∞0xmxn+an=πam+1−nnsin[(m+1)π/n], 0<m+1<n
∫a0dxa2−x2−−−−−−√=π2
∫a0a2−x2−−−−−−√dx=πa24
∫a0xm(an−xn)pdx=am+1+np Γ[(m+1)/n] Γ(p+1)nΓ[(m+1)/n+p+1]
∫π/20sin2xdx=∫π/20cos2xdx=π4
∫∞0sin(px)xdx=⎧⎩⎨⎪⎪π/2 0−π/2p>0p=0p<0
∫∞0sin2pxx2=πp2
∫∞01−cos(px)x2dx=πp2
∫∞0cos(px)−cos(qx)xdx=lnqp
∫∞0cos(px)−cos(qx)x2dx=π(q−p)2
∫2π0dxa+bsinx=2πa2−b2−−−−−−√
∫2π0dxa+bcos(x)=2πa2−b2−−−−−−√
∫∞0sinax2dx=∫∞0cos(ax2)dx=12π2a−−−√
∫∞0sinxx√dx=∫∞0cosxx√dx=π2−−√
∫∞0sin3xx3dx=3π8
∫∞0sin4xx4dx=π3
∫∞0tanxxdx=π2
∫π/20dxa+bcosx=arccos(b/a)a2−b2−−−−−−√

Advanced formulas

∫π0sin(mx)⋅sin(nx)dx={0π/2m,n integers and m≠nm,n integers and m=n
∫π0cos(mx)⋅cos(nx)dx={0π/2m,n integers and m≠nm,n integers and m=n
∫π0sin(mx)⋅cos(nx)dx={02m/(m2−n2)m,n integers and m+n oddm,n integers and m+n even
∫π/20sin2mxdx=∫π/20cos2mxdx=1⋅3⋅5…2m−12⋅4⋅6…2mπ2
∫π/20sin2m+1xdx=∫π/20cos2m+1xdx=2⋅4⋅6…2m1⋅3⋅5…2m+1
∫π0sin2p−1xcos2q−1xdx=Γ(p)Γq2Γ(p+q)
∫∞0sin(px)⋅cos(qx)xdx=⎧⎩⎨⎪⎪ 0π/2π/4p>q>00<p<qp=q>0
∫∞0sin(px)⋅sin(qx)x2dx={πp/2πq/20<p≤qp≥q>0
∫∞0cos(mx)x2+a2dx=π2ae−ma
∫∞0xsin(mx)x2+a2dx=π2e−ma
∫∞0sin(mx)x(x2+a2)dx=π2a2(1−e−ma)
∫2π0dx(a+bsinx)2=∫2π0dx(a+bcosx)2=2πa(a2−b2)3/2
∫2π0dx1−2acosx+a2=2π1−a2,  0<a<1
∫π0xsinxdx1−2acosx+a2=⎧⎩⎨⎪⎪⎪⎪πaln(1+a)πln(1+1a)|a|<1|a|>1
∫π0cos(mx)dx1−2acosx+a2=πam1−a2,  a2<1
∫∞0sin(axn)dx=1na1/nΓ(1/n)sinπ2n,  n>1
∫∞0cos(axn)dx=1na1/nΓ(1/n)cosπ2n,  n>1
∫∞0sinxxpdx=π2Γ(p)sin(pπ/2),  0<p<1
∫∞0cosxxpdx=π2Γ(p)cos(pπ/2),  0<p<1
∫∞0sin(ax2)cos(2bx)dx=12π2a−−−√(cosb2a−sinb2a)
∫∞0cos(ax2)cos(2bx)dx=12π2a−−−√(cosb2a+sinb2a)
∫∞0dx1+tanmxdx=π4
∫∞0e−axcosbxdx=aa2+b2
∫∞0e−axsinbxdx=ba2+b2
∫∞0e−axsinbxxdx=arctanba
∫∞0e−ax−e−bxxdx=lnba
∫∞0e−ax2dx=12πa−−√
∫∞0e−ax2cosbxdx=12πa−−√e−b24a
∫∞−∞e−(ax2+bx+c)dx=π2−−√eb2−4ac4a
∫∞0xne−axdx=Γ(n+1)an+1
∫∞0xme−ax2dx=Γ(m+12)2a(m+1)/2
∫∞0e−(ax2+b/x2)dx=12πa−−√e−2ab√
∫∞0xdxex−1=π26
∫∞0xn−1ex−1dx=Γ(n)(11n+12n+13n+⋯)
∫∞0xdxex+1=π212
∫∞0xn−1ex+1dx=Γ(n)(11n−12n+13n−⋯)
∫∞0sinmxe2πx−1dx=14cothm2−12m
∫∞0(11+x−e−x)dxx=γ
∫∞0e−x2−e−xxdx=12γ
∫∞0(1ex−1−e−xx)dx=γ
∫∞0e−ax−e−bxxsec(px)dx=12ln(b2+p2a2+p2)
∫∞0e−ax−e−bxxcsc(px)dx=arctanbp−arctanap
∫∞0e−ax(1−cosx)x2dx=arccota−a2ln(a2+1)
∫10xm(lnx)ndx=(−1)nn!(m+1)n+1,m>−1,n=0,1,2,…
∫10lnx1+xdx=−π212
∫10lnx1−xdx=−π26
∫10ln(1+x)xdx=π212
∫10ln(1−x)xdx=−π26
∫10lnxln(1+x)dx=2−2ln2−π212
∫10lnxln(1−x)dx=2−π26
∫∞0xp−1lnx1+xdx=−π2csc(pπ)cot(pπ),0<p<1
∫10xm−xnlnxdx=lnm+1n+1
∫∞0e−xlnxdx=−γ
∫∞0e−x2lnxdx=−π√4(γ+2ln2)
∫∞0ln(ex+1ex−1)dx=π24
∫π/20ln(sinx)dx=∫π/20ln(cosx)dx=−π2ln2
∫π/20(ln(sinx))2dx=∫π/20(ln(cosx))2dx=π2(ln2)2+π324
∫π0xln(sinx)dx=−π22ln2
∫π/20sinxln(sinx)dx=ln2−1
∫2π0ln(a+bsinx)dx=∫2π0ln(a+bcosx)dx=2πln(a+a2−b2−−−−−−√)
∫π0ln(a+bcosx)dx=πln(a+a2−b2−−−−−−√2)
∫π0ln(a2−2abcosx+b2)dx={2πlna2πlnba≥b>0b≥a>0
∫π/40ln(1+tanx)dx=π8ln2
∫π20secxln(1+bcosx1+acosx)dx=12(arccos2a−arccos2b)

 

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