गणित में निश्चित समाकल ∫ a b f ( x ) d x {\displaystyle \int _{a}^{b}f(x)\,dx} xy-समतल के ग्राफ f, x-अक्ष, तथा x = a और x = b रेखाओं से घिरे हुए क्षेत्र के क्षेत्रफल के बराबर होता है। (x-अक्ष के ऊपर का क्षेत्रफल धनात्मक लेते हैं जबकि x-अक्ष के नीचे का क्षेत्र ऋणात्मक) ∫ 0 ∞ d x x 2 + a 2 = π 2 a {\displaystyle \int _{0}^{\infty }{\frac {dx}{x^{2}+a^{2}}}={\frac {\pi }{2a}}} ∫ 0 a x m d x x n + a n = π a m − n + 1 n sin [ ( m + 1 ) π / n ) ] {\displaystyle \int _{0}^{a}{\frac {x^{m}dx}{x^{n}+a^{n}}}={\frac {\pi a^{m-n+1}}{n\sin[(m+1)\pi /n)]}}} ∫ 0 ∞ x p − 1 d x 1 + x = π sin p π 0 < p < 1 {\displaystyle \int _{0}^{\infty }{\frac {x^{p-1}dx}{1+x}}={\frac {\pi }{\sin p\pi }}\ \,0<p<1} ∫ 0 ∞ x m d x 1 + 2 x cos β + x 2 = π sin ( m π ) sin ( m β ) sin β {\displaystyle \int _{0}^{\infty }{\frac {x^{m}dx}{1+2x\cos \beta +x^{2}}}={\frac {\pi }{\sin(m\pi )}}{\frac {\sin(m\beta )}{\sin \beta }}} ∫ 0 ∞ d x a 2 − x 2 = π 2 {\displaystyle \int _{0}^{\infty }{\frac {dx}{\sqrt {a^{2}-x^{2}}}}={\frac {\pi }{2}}} ∫ 0 a a 2 − x 2 d x = π a 2 4 {\displaystyle \int _{0}^{a}{\sqrt {a^{2}-x^{2}}}\,dx={\frac {\pi a^{2}}{4}}} ∫ 0 a x m ( a n − x n ) p d x = a m + 1 + n p Γ [ ( m + 1 ) / n ] Γ ( p + 1 ) n Γ [ ( ( m + 1 ) / n ) + p + 1 ] {\displaystyle \int _{0}^{a}x^{m}(a^{n}-x^{n})^{p}\,dx={\frac {a^{m+1+np}\Gamma [(m+1)/n]\Gamma (p+1)}{n\Gamma [((m+1)/n)+p+1]}}} ∫ 0 ∞ x m d x ( x n + a n ) r = ( − 1 ) r − 1 π a m + 1 − n r Γ [ ( m + 1 ) / n ] n sin [ ( m + 1 ) π / n ] ( r − 1 ) ! Γ [ ( m + 1 ) / n − r + 1 ] 0 < m + 1 < n r {\displaystyle \int _{0}^{\infty }{\frac {x^{m}\,dx}{({x^{n}+a^{n})}^{r}}}={\frac {(-1)^{r-1}\pi a^{m+1-nr}\Gamma [(m+1)/n]}{n\sin[(m+1)\pi /n](r-1)!\Gamma [(m+1)/n-r+1]}}\ \,0<m+1<nr} ∫ 0 π sin m x sin n x d x = { 0 if m ≠ n π / 2 if m = n m , n integers {\displaystyle \int _{0}^{\pi }\sin mx\sin nx\,dx={\begin{cases}0&{\text{if }}m\neq n\\\pi /2&{\text{if }}m=n\end{cases}}\ \ m,n{\text{ integers}}} ∫ 0 π cos m x cos n x d x = { 0 if m ≠ n π / 2 if m = n m , n integers {\displaystyle \int _{0}^{\pi }\cos mx\cos nxdx={\begin{cases}0&{\text{if }}m\neq n\\\pi /2&{\text{if }}m=n\end{cases}}\ \ m,n{\text{ integers}}} ∫ 0 π sin m x cos n x d x = { 0 if m + n even 2 m m 2 − n 2 if m + n odd m , n integers . {\displaystyle \int _{0}^{\pi }\sin mx\cos nx\,dx={\begin{cases}0&{\text{if }}m+n{\text{ even}}\\{\frac {2m}{m^{2}-n^{2}}}&{\text{if }}m+n{\text{ odd}}\end{cases}}\ \ m,n{\text{ integers}}.} ∫ 0 π / 2 sin 2 x d x = ∫ 0 π / 2 cos 2 x d x = π / 4 {\displaystyle \int _{0}^{\pi /2}\sin ^{2}x\,dx=\int _{0}^{\pi /2}\cos ^{2}x\,dx=\pi /4} ∫ 0 π / 2 sin 2 m x d x = ∫ 0 π / 2 cos 2 m x d x = 1 × 3 × 5 × ⋯ × ( 2 m − 1 ) 2 × 4 × 6 × ⋯ × 2 m π 2 m = 1 , 2 , 3 , … {\displaystyle \int _{0}^{\pi /2}\sin ^{2m}x\,dx=\int _{0}^{\pi /2}\cos ^{2m}x\,dx={\frac {1\times 3\times 5\times \cdots \times (2m-1)}{2\times 4\times 6\times \cdots \times 2m}}{\frac {\pi }{2}}\ \ m=1,2,3,\ldots } ∫ 0 π / 2 sin 2 m + 1 x d x = ∫ 0 π / 2 cos 2 m + 1 x d x = 2 × 4 × 6 × ⋯ × 2 m 1 × 3 × 5 × ⋯ × ( 2 m − 1 ) m = 1 , 2 , 3 , … {\displaystyle \int _{0}^{\pi /2}\sin ^{2m+1}x\,dx=\int _{0}^{\pi /2}\cos ^{2m+1}x\,dx={\frac {2\times 4\times 6\times \cdots \times 2m}{1\times 3\times 5\times \cdots \times (2m-1)}}\ \ m=1,2,3,\ldots } ∫ 0 π / 2 sin 2 p − 1 cos 2 q − 1 x d x = Γ ( p ) Γ ( q ) 2 Γ ( p + q ) {\displaystyle \int _{0}^{\pi /2}\sin ^{2p-1}\cos ^{2q-1}x\,dx={\frac {\Gamma (p)\Gamma (q)}{2\Gamma (p+q)}}} ∫ 0 ∞ sin p x x d x = { π / 2 if p > 0 0 if p = 0 − π / 2 if p < 0 {\displaystyle \int _{0}^{\infty }{\frac {\sin px}{x}}\,dx={\begin{cases}\pi /2&{\text{if }}p>0\\0&{\text{if }}p=0\\-\pi /2&{\text{if }}p<0\end{cases}}} ∫ 0 ∞ sin p x cos q x x d x = { 0 if p > q > 0 π / 2 if 0 < p < q π / 4 if p = q > 0 {\displaystyle \int _{0}^{\infty }{\frac {\sin px\cos qx}{x}}\ dx={\begin{cases}0&{\text{ if }}p>q>0\\\pi /2&{\text{ if }}0<p<q\\\pi /4&{\text{ if }}p=q>0\end{cases}}} ∫ 0 ∞ sin p x sin q x x 2 d x = { π p / 2 if 0 < p ≤ q π q / 2 if 0 < q ≤ p {\displaystyle \int _{0}^{\infty }{\frac {\sin px\sin qx}{x^{2}}}\ dx={\begin{cases}\pi p/2&{\text{ if }}0<p\leq q\\\pi q/2&{\text{ if }}0<q\leq p\end{cases}}} ∫ 0 ∞ sin 2 p x x 2 d x = π p 2 {\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}px}{x^{2}}}\ dx={\frac {\pi p}{2}}} ∫ 0 ∞ 1 − cos p x x 2 d x = π p 2 {\displaystyle \int _{0}^{\infty }{\frac {1-\cos px}{x^{2}}}\ dx={\frac {\pi p}{2}}} ∫ 0 ∞ cos p x − cos q x x d x = ln q p {\displaystyle \int _{0}^{\infty }{\frac {\cos px-\cos qx}{x}}\ dx=\ln {\frac {q}{p}}} ∫ 0 ∞ cos p x − cos q x x 2 d x = π ( q − p ) 2 {\displaystyle \int _{0}^{\infty }{\frac {\cos px-\cos qx}{x^{2}}}\ dx={\frac {\pi (q-p)}{2}}} ∫ 0 ∞ cos m x x 2 + a 2 d x = π 2 a e − m a {\displaystyle \int _{0}^{\infty }{\frac {\cos mx}{x^{2}+a^{2}}}\ dx={\frac {\pi }{2a}}e^{-ma}} ∫ 0 ∞ x sin m x x 2 + a 2 d x = π 2 e − m a {\displaystyle \int _{0}^{\infty }{\frac {x\sin mx}{x^{2}+a^{2}}}\ dx={\frac {\pi }{2}}e^{-ma}} ∫ 0 ∞ sin m x x ( x 2 + a 2 ) d x = π 2 a 2 ( 1 − e − m a ) {\displaystyle \int _{0}^{\infty }{\frac {\sin mx}{x(x^{2}+a^{2})}}\ dx={\frac {\pi }{2a^{2}}}(1-e^{-ma})} ∫ 0 2 π d x a + b sin x = 2 π a 2 − b 2 {\displaystyle \int _{0}^{2\pi }{\frac {dx}{a+b\sin x}}={\frac {2\pi }{\sqrt {a^{2}-b^{2}}}}} ∫ 0 2 π d x a + b cos x = 2 π a 2 − b 2 {\displaystyle \int _{0}^{2\pi }{\frac {dx}{a+b\cos x}}={\frac {2\pi }{\sqrt {a^{2}-b^{2}}}}} ∫ 0 π 2 d x a + b cos x = cos − 1 ( b / a ) a 2 − b 2 {\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {dx}{a+b\cos x}}={\frac {\cos ^{-1}(b/a)}{\sqrt {a^{2}-b^{2}}}}} ∫ 0 2 π d x ( a + b sin x ) 2 = ∫ 0 2 π d x ( a + b cos x ) 2 = 2 π a ( a 2 − b 2 ) 3 / 2 {\displaystyle \int _{0}^{2\pi }{\frac {dx}{(a+b\sin x)^{2}}}=\int _{0}^{2\pi }{\frac {dx}{(a+b\cos x)^{2}}}={\frac {2\pi a}{(a^{2}-b^{2})^{3/2}}}} ∫ 0 2 π d x 1 − 2 a cos x + a 2 = 2 π 1 − a 2 0 < a < 1 {\displaystyle \int _{0}^{2\pi }{\frac {dx}{1-2a\cos x+a^{2}}}={\frac {2\pi }{1-a^{2}}}\ \ \,\ 0<a<1} ∫ 0 π x sin x d x 1 − 2 a cos x + a 2 = { π a ln ( 1 + a ) if | a | < 1 π ln ( 1 + 1 / a ) if | a | > 1 {\displaystyle \int _{0}^{\pi }{\frac {x\sin x\ dx}{1-2a\cos x+a^{2}}}={\begin{cases}{\frac {\pi }{a}}\ln(1+a)&{\text{if }}|a|<1\\\pi \ln(1+1/a)&{\text{if }}|a|>1\end{cases}}} ∫ 0 π cos m x d x 1 − 2 a cos x + a 2 = π a m 1 − a 2 , a 2 < 1 , m = 0 , 1 , 2 , … {\displaystyle \int _{0}^{\pi }{\frac {\cos mx\ dx}{1-2a\cos x+a^{2}}}={\frac {\pi a^{m}}{1-a^{2}}}\quad ,a^{2}<1,\ m=0,1,2,\dots } ∫ 0 ∞ sin a x 2 d x = ∫ 0 ∞ cos a x 2 = 1 2 π 2 a {\displaystyle \int _{0}^{\infty }\sin ax^{2}\ dx=\int _{0}^{\infty }\cos ax^{2}={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}} ∫ 0 ∞ sin a x n = 1 n a 1 / n Γ ( 1 / n ) sin π 2 n , n > 1 {\displaystyle \int _{0}^{\infty }\sin ax^{n}={\frac {1}{na^{1/n}}}\Gamma (1/n)\sin {\frac {\pi }{2n}}\quad ,n>1} ∫ 0 ∞ cos a x n = 1 n a 1 / n Γ ( 1 / n ) cos π 2 n , n > 1 {\displaystyle \int _{0}^{\infty }\cos ax^{n}={\frac {1}{na^{1/n}}}\Gamma (1/n)\cos {\frac {\pi }{2n}}\quad ,n>1} ∫ 0 ∞ sin x x d x = ∫ 0 ∞ cos x x d x = π 2 {\displaystyle \int _{0}^{\infty }{\frac {\sin x}{\sqrt {x}}}\ dx=\int _{0}^{\infty }{\frac {\cos x}{\sqrt {x}}}\ dx={\sqrt {\frac {\pi }{2}}}} ∫ 0 ∞ sin x x p d x = π 2 Γ ( p ) sin ( p π / 2 ) , 0 < p < 1 {\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x^{p}}}\ dx={\frac {\pi }{2\Gamma (p)\sin(p\pi /2)}},\quad 0<p<1} ∫ 0 ∞ cos x x p d x = π 2 Γ ( p ) cos ( p π / 2 ) , 0 < p < 1 {\displaystyle \int _{0}^{\infty }{\frac {\cos x}{x^{p}}}\ dx={\frac {\pi }{2\Gamma (p)\cos(p\pi /2)}},\quad 0<p<1} ∫ 0 ∞ sin a x 2 cos 2 b x d x = 1 2 π 2 a ( cos b 2 a − sin b 2 a ) {\displaystyle \int _{0}^{\infty }\sin ax^{2}\cos 2bx\ dx={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}(\cos {\frac {b^{2}}{a}}-\sin {\frac {b^{2}}{a}})} ∫ 0 ∞ cos a x 2 cos 2 b x d x = 1 2 π 2 a ( cos b 2 a + sin b 2 a ) {\displaystyle \int _{0}^{\infty }\cos ax^{2}\cos 2bx\ dx={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}(\cos {\frac {b^{2}}{a}}+\sin {\frac {b^{2}}{a}})} ∫ 0 ∞ e − a x cos b x d x = a a 2 + b 2 {\displaystyle \int _{0}^{\infty }e^{-ax}\cos bx\,dx={\frac {a}{a^{2}+b^{2}}}} ∫ 0 ∞ e − a x sin b x d x = b a 2 + b 2 {\displaystyle \int _{0}^{\infty }e^{-ax}\sin bx\,dx={\frac {b}{a^{2}+b^{2}}}} ∫ 0 ∞ e − a x sin b x x d x = tan − 1 b a {\displaystyle \int _{0}^{\infty }{\frac {{}e^{-ax}\sin bx}{x}}\,dx=\tan ^{-1}{\frac {b}{a}}} ∫ 0 ∞ e − a x − e − b x x d x = ln b a {\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x}}\,dx=\ln {\frac {b}{a}}} ∫ 0 ∞ e − a x 2 d x = 1 2 π a {\displaystyle \int _{0}^{\infty }{e^{-ax^{2}}}\,dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}} ∫ 0 ∞ e − a x 2 cos b x d x = 1 2 π a e − b 2 / 4 a {\displaystyle \int _{0}^{\infty }{e^{-ax^{2}}}\cos bx\,dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}e^{-b^{2}/4a}} ∫ 0 ∞ e − ( a x 2 + b x + c ) d x = 1 2 π a e ( b 2 − 4 a c ) / 4 a erfc b 2 a , where erfc ( p ) = 2 π ∫ p ∞ e − x 2 d x {\displaystyle \int _{0}^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}e^{(b^{2}-4ac)/4a}\ \operatorname {erfc} {\frac {b}{2{\sqrt {a}}}},{\text{ where }}\operatorname {erfc} (p)={\frac {2}{\sqrt {\pi }}}\int _{p}^{\infty }e^{-x^{2}}\,dx} ∫ − ∞ + ∞ e − ( a x 2 + b x + c ) d x = π a e ( b 2 − 4 a c ) / 4 a {\displaystyle \int _{-\infty }^{+\infty }e^{-(ax^{2}+bx+c)}\ dx={\sqrt {\frac {\pi }{a}}}e^{(b^{2}-4ac)/4a}} ∫ 0 ∞ x n e − a x d x = Γ ( n + 1 ) a n + 1 {\displaystyle \int _{0}^{\infty }x^{n}e^{-ax}\ dx={\frac {\Gamma (n+1)}{a^{n+1}}}} ∫ 0 ∞ x m e − a x 2 d x = Γ [ ( m + 1 ) / 2 ] 2 a ( m + 1 ) / 2 {\displaystyle \int _{0}^{\infty }x^{m}e^{-ax^{2}}\ dx={\frac {\Gamma [(m+1)/2]}{2a^{(m+1)/2}}}} ∫ 0 ∞ e − a x 2 − b / x 2 d x = 1 2 π a e − 2 a b {\displaystyle \int _{0}^{\infty }e^{-ax^{2}-b/x^{2}}\ dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}e^{-2{\sqrt {ab}}}} ∫ 0 ∞ x e x − 1 d x = ζ ( 2 ) = π 2 6 {\displaystyle \int _{0}^{\infty }{\frac {x}{e^{x}-1}}\ dx=\zeta (2)={\frac {\pi ^{2}}{6}}} ∫ 0 ∞ x n − 1 e x − 1 d x = Γ ( n ) ζ ( n ) {\displaystyle \int _{0}^{\infty }{\frac {x^{n-1}}{e^{x}-1}}\ dx=\Gamma (n)\zeta (n)} ∫ 0 ∞ x e x + 1 d x = 1 1 2 − 1 2 2 + 1 3 2 − 1 4 2 + ⋯ = π 2 12 {\displaystyle \int _{0}^{\infty }{\frac {x}{e^{x}+1}}\ dx={\frac {1}{1^{2}}}-{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}-{\frac {1}{4^{2}}}+\dots ={\frac {\pi ^{2}}{12}}} ∫ 0 ∞ sin m x e 2 π x − 1 d x = 1 4 coth m 2 − 1 2 m {\displaystyle \int _{0}^{\infty }{\frac {\sin mx}{e^{2\pi x}-1}}\ dx={\frac {1}{4}}\coth {\frac {m}{2}}-{\frac {1}{2m}}} ∫ 0 ∞ ( 1 1 + x − e − x ) d x x = γ {\displaystyle \int _{0}^{\infty }({\frac {1}{1+x}}-e^{-x}){\frac {dx}{x}}=\gamma } ∫ 0 ∞ e − x 2 − e − x x d x = γ 2 {\displaystyle \int _{0}^{\infty }{\frac {e^{-x^{2}}-e^{-x}}{x}}\ dx={\frac {\gamma }{2}}} ∫ 0 ∞ ( 1 e x − 1 − e − x x ) d x = γ {\displaystyle \int _{0}^{\infty }({\frac {1}{e^{x}-1}}-{\frac {e^{-x}}{x}})dx=\gamma } ∫ 0 ∞ e − a x − e − b x x sec p x d x = 1 2 ln b 2 + p 2 a 2 + p 2 {\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x\sec px}}\ dx={\frac {1}{2}}\ln {\frac {b^{2}+p^{2}}{a^{2}+p^{2}}}} ∫ 0 ∞ e − a x − e − b x x csc p x d x = tan − 1 b p − tan − 1 a p {\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x\csc px}}\ dx=\tan ^{-1}{\frac {b}{p}}-\tan ^{-1}{\frac {a}{p}}} ∫ 0 ∞ e − a x ( 1 − cos x ) x 2 d x = cot − 1 a − a 2 ln ( a 2 + 1 ) {\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}(1-\cos x)}{x^{2}}}\ dx=\cot ^{-1}a-{\frac {a}{2}}\ln(a^{2}+1)} ∫ − ∞ ∞ e − x 2 d x = π {\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}} ∫ − ∞ ∞ x 2 ( n + 1 ) e − x 2 / 2 d x = ( 2 n + 1 ) ! 2 n n ! 2 π n = 0 , 1 , 2 , … {\displaystyle \int _{-\infty }^{\infty }x^{2(n+1)}e^{-x^{2}/2}\,dx={\frac {(2n+1)!}{2^{n}n!}}{\sqrt {2\pi }}\quad n=0,1,2,\ldots } ∫ 0 1 x m ( ln x ) n d x = ( − 1 ) n n ! ( m + 1 ) n + 1 m > − 1 , n = 0 , 1 , 2 , … {\displaystyle \int _{0}^{1}x^{m}(\ln x)^{n}\,dx={\frac {(-1)^{n}n!}{(m+1)^{n+1}}}\quad m>-1,n=0,1,2,\ldots } ∫ 0 1 ln x 1 + x d x = − π 2 12 {\displaystyle \int _{0}^{1}{\frac {\ln x}{1+x}}\,dx=-{\frac {\pi ^{2}}{12}}} ∫ 0 1 ln x 1 − x d x = − π 2 6 {\displaystyle \int _{0}^{1}{\frac {\ln x}{1-x}}\,dx=-{\frac {\pi ^{2}}{6}}} ∫ 0 1 ln ( 1 + x ) x d x = π 2 12 {\displaystyle \int _{0}^{1}{\frac {\ln(1+x)}{x}}\,dx={\frac {\pi ^{2}}{12}}} ∫ 0 1 ln ( 1 − x ) x d x = − π 2 6 {\displaystyle \int _{0}^{1}{\frac {\ln(1-x)}{x}}\,dx=-{\frac {\pi ^{2}}{6}}} ∫ 0 ∞ sin a x sinh b x d x = π 2 b tanh a π 2 b {\displaystyle \int _{0}^{\infty }{\frac {\sin ax}{\sinh bx}}\ dx={\frac {\pi }{2b}}\tanh {\frac {a\pi }{2b}}} ∫ 0 ∞ cos a x cosh b x d x = π 2 b 1 cosh a π 2 b {\displaystyle \int _{0}^{\infty }{\frac {\cos ax}{\cosh bx}}\ dx={\frac {\pi }{2b}}{\frac {1}{\cosh {\frac {a\pi }{2b}}}}} ∫ 0 ∞ x sinh a x d x = π 2 4 a 2 {\displaystyle \int _{0}^{\infty }{\frac {x}{\sinh ax}}\ dx={\frac {\pi ^{2}}{4a^{2}}}} ∫ 0 ∞ f ( a x ) − f ( b x ) x d x = [ f ( 0 ) − f ( ∞ ) ] ln b a {\displaystyle \int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\ dx=[{f(0)-f(\infty )}]\ln {\frac {b}{a}}} ∫ − a a ( a + x ) m − 1 ( a − x ) n − 1 d x = ( 2 a ) m + n − 1 Γ ( m ) Γ ( n ) Γ ( m + n ) {\displaystyle \int _{-a}^{a}(a+x)^{m-1}(a-x)^{n-1}\ dx=(2a)^{m+n-1}{\frac {\Gamma (m)\Gamma (n)}{\Gamma (m+n)}}} समाकलों की सूची (List of integrals) गामा गलन (Gamma function) सीमाओं की सूची (List of limits) ↑Murray R. Spiegel, Seymour Lipschutz, John Liu (2009). Mathematical handbook of formulas and tables (3rd ed. संस्करण). McGraw-Hill. आई॰ऍस॰बी॰ऍन॰ 978-0071548557. मूल से 2 अक्तूबर 2019 को पुरालेखित. अभिगमन तिथि 23 अक्तूबर 2019.सीएस1 रखरखाव: एक से अधिक नाम: authors list (link) सीएस1 रखरखाव: फालतू पाठ (link)↑Zwillinger, Daniel (2003). CRC standard mathematical tables and formulae (32nd ed. संस्करण). CRC Press. आई॰ऍस॰बी॰ऍन॰ 978-1439835487. मूल से 2 अक्तूबर 2019 को पुरालेखित. अभिगमन तिथि 23 अक्तूबर 2019.सीएस1 रखरखाव: फालतू पाठ (link)↑Abramowitz, Milton; Stegun, Irene A. (1965). Handbook of mathematical functions with formulas, graphs, and mathematical tables (Unabridged and unaltered republ. [der Ausg.] 1964, 5. Dover printing संस्करण). U.S. Govt. Print. Off. आई॰ऍस॰बी॰ऍन॰ 978-0486612720.Wikiwand in your browser!Seamless Wikipedia browsing. 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Every time you click a link to Wikipedia, Wiktionary or Wikiquote in your browser's search results, it will show the modern Wikiwand interface.Wikiwand extension is a five stars, simple, with minimum permission required to keep your browsing private, safe and transparent.