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A new editor on Wikipedia ! Will try my level best to make Wikipedia better!
VSOian Name Joshua
Born (2006-05-07 ) May 7, 2006 (age 18) Name in real life Vinayak Singh Nationality Indian Country India Current location Kolkata Time zone Indian Standard Time Race Asian Height 5'8" Weight 116 lbs. Eyes Black and Hypnotizing Blood type Type B +ve Sexuality Heterosexual , Straight IQ 136 Personality type Resilient, protective, controlling, ambitious, ruthless, deceptive, strategic, authoritative, hardened by experiences, survival-driven, class-conscious, influential, and authoritative Marital status Unmarried- Marriage is the fastest and easiest way to start disliking the person you like! Children None, Doesn't want to have any also! Siblings Abhishek Singh (b.1988) Parents Uma Dhar (my mother) and Rajiv Singh (my papa) Pets 1 dog (Happy) Occupation Author High school Army Public School, Ballygunge Hobbies Writing, eating, sleeping, travelling Religion Hindu Politics Bharatiya Janata Party Shows Friends , The Big Bang Theory Influenced by Sakshi Goenka in- Ek Hasina Thi (TV series) Books Harry Potter Joined January 6, 2024
This user lives in India
This user has Knight
This user has a May birthday
♂ This contributor to Wikipedia is male female !
This user is proud to have studied in a CBSE India .
This user is a cat .
This user is proud to be anIndian
This user is proud to be anAsian
This user is a scary Ghost
This user likes waves.
∑
k
=
1
∞
x
k
k
!
{\displaystyle \sum _{k=1}^{\infty }{x^{k} \over k!}}
This user loves problem solving .
∫
e
a
x
d
x
=
1
a
e
a
x
+
C
{\displaystyle \int e^{ax}\,dx={\frac {1}{a}}e^{ax}+C}
∫
f
′
(
x
)
e
f
(
x
)
d
x
=
e
f
(
x
)
+
C
{\displaystyle \int f'(x)e^{f(x)}\,dx=e^{f(x)}+C}
∫
a
x
d
x
=
a
x
ln
a
+
C
{\displaystyle \int a^{x}\,dx={\frac {a^{x}}{\ln a}}+C}
∫
e
x
(
f
(
x
)
+
f
′
(
x
)
)
d
x
=
e
x
f
(
x
)
+
C
{\displaystyle \int {e^{x}\left(f\left(x\right)+f'\left(x\right)\right)\,dx}=e^{x}f\left(x\right)+C}
∫
e
x
(
f
(
x
)
−
(
−
1
)
n
d
n
f
(
x
)
d
x
n
)
d
x
=
e
x
∑
k
=
1
n
(
−
1
)
k
−
1
d
k
−
1
f
(
x
)
d
x
k
−
1
+
C
{\displaystyle \int {e^{x}\left(f\left(x\right)-\left(-1\right)^{n}{\frac {d^{n}f\left(x\right)}{dx^{n}}}\right)\,dx}=e^{x}\sum _{k=1}^{n}{\left(-1\right)^{k-1}{\frac {d^{k-1}f\left(x\right)}{dx^{k-1}}}}+C}
(if
n
{\displaystyle n}
∫
e
−
x
(
f
(
x
)
−
d
n
f
(
x
)
d
x
n
)
d
x
=
−
e
−
x
∑
k
=
1
n
d
k
−
1
f
(
x
)
d
x
k
−
1
+
C
{\displaystyle \int {e^{-x}\left(f\left(x\right)-{\frac {d^{n}f\left(x\right)}{dx^{n}}}\right)\,dx}=-e^{-x}\sum _{k=1}^{n}{\frac {d^{k-1}f\left(x\right)}{dx^{k-1}}}+C}
(if
n
{\displaystyle n}
∫
0
∞
x
e
−
x
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{\sqrt {x}}\,e^{-x}\,dx={\frac {1}{2}}{\sqrt {\pi }}}
Gamma function )
∫
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}}}}
a > 0Gaussian integral )
∫
0
∞
x
2
e
−
a
x
2
d
x
=
1
4
π
a
3
{\displaystyle \int _{0}^{\infty }{x^{2}e^{-ax^{2}}\,dx}={\frac {1}{4}}{\sqrt {\frac {\pi }{a^{3}}}}}
a > 0
for a > 0n !! is the double factorial .
∫
0
∞
x
3
e
−
a
x
2
d
x
=
1
2
a
2
{\displaystyle \int _{0}^{\infty }{x^{3}e^{-ax^{2}}\,dx}={\frac {1}{2a^{2}}}}
a > 0
∫
0
∞
x
2
n
+
1
e
−
a
x
2
d
x
=
n
a
∫
0
∞
x
2
n
−
1
e
−
a
x
2
d
x
=
n
!
2
a
n
+
1
{\displaystyle \int _{0}^{\infty }x^{2n+1}e^{-ax^{2}}\,dx={\frac {n}{a}}\int _{0}^{\infty }x^{2n-1}e^{-ax^{2}}\,dx={\frac {n!}{2a^{n+1}}}}
for a > 0n = 0, 1, 2, ....
∫
0
∞
x
e
x
−
1
d
x
=
π
2
6
{\displaystyle \int _{0}^{\infty }{\frac {x}{e^{x}-1}}\,dx={\frac {\pi ^{2}}{6}}}
Bernoulli number )
∫
0
∞
x
2
e
x
−
1
d
x
=
2
ζ
(
3
)
≈
2.40
{\displaystyle \int _{0}^{\infty }{\frac {x^{2}}{e^{x}-1}}\,dx=2\zeta (3)\approx 2.40}
∫
0
∞
x
3
e
x
−
1
d
x
=
π
4
15
{\displaystyle \int _{0}^{\infty }{\frac {x^{3}}{e^{x}-1}}\,dx={\frac {\pi ^{4}}{15}}}
∫
0
∞
sin
x
x
d
x
=
π
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin {x}}{x}}\,dx={\frac {\pi }{2}}}
sinc function and the Dirichlet integral )
∫
0
∞
sin
2
x
x
2
d
x
=
π
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}{x}}{x^{2}}}\,dx={\frac {\pi }{2}}}
∫
0
π
2
sin
n
x
d
x
=
∫
0
π
2
cos
n
x
d
x
=
(
n
−
1
)
!
!
n
!
!
×
{
1
if
n
is odd
π
2
if
n
is even.
{\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}x\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{n}x\,dx={\frac {(n-1)!!}{n!!}}\times {\begin{cases}1&{\text{if }}n{\text{ is odd}}\\{\frac {\pi }{2}}&{\text{if }}n{\text{ is even.}}\end{cases}}}
(if n double factorial ).
∫
−
π
π
cos
(
α
x
)
cos
n
(
β
x
)
d
x
=
{
2
π
2
n
(
n
m
)
|
α
|
=
|
β
(
2
m
−
n
)
|
0
otherwise
{\displaystyle \int _{-\pi }^{\pi }\cos(\alpha x)\cos ^{n}(\beta x)dx={\begin{cases}{\frac {2\pi }{2^{n}}}{\binom {n}{m}}&|\alpha |=|\beta (2m-n)|\\0&{\text{otherwise}}\end{cases}}}
(for α , β , m , n β ≠ 0m , n ≥ 0Binomial coefficient )
∫
−
t
t
sin
m
(
α
x
)
cos
n
(
β
x
)
d
x
=
0
{\displaystyle \int _{-t}^{t}\sin ^{m}(\alpha x)\cos ^{n}(\beta x)dx=0}
(for α , β n m an odd, positive integer; since the integrand is odd )
∫
−
∞
∞
e
−
(
a
x
2
+
b
x
+
c
)
d
x
=
π
a
exp
[
b
2
−
4
a
c
4
a
]
{\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\sqrt {\frac {\pi }{a}}}\exp \left[{\frac {b^{2}-4ac}{4a}}\right]}
(where exp[u ] is the exponential function e u a > 0
∫
0
∞
x
z
−
1
e
−
x
d
x
=
Γ
(
z
)
{\displaystyle \int _{0}^{\infty }x^{z-1}\,e^{-x}\,dx=\Gamma (z)}
(where
Γ
(
z
)
{\displaystyle \Gamma (z)}
Gamma function )
∫
0
1
(
ln
1
x
)
p
d
x
=
Γ
(
p
+
1
)
{\displaystyle \int _{0}^{1}\left(\ln {\frac {1}{x}}\right)^{p}\,dx=\Gamma (p+1)}
∫
0
1
x
α
−
1
(
1
−
x
)
β
−
1
d
x
=
Γ
(
α
)
Γ
(
β
)
Γ
(
α
+
β
)
{\displaystyle \int _{0}^{1}x^{\alpha -1}(1-x)^{\beta -1}dx={\frac {\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}}}
(for Re(α ) > 0 and Re(β ) > 0 , see Beta function )
∫
0
2
π
e
x
cos
θ
d
θ
=
2
π
I
0
(
x
)
{\displaystyle \int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)}
I 0 (x )Bessel function of the first kind)
∫
0
2
π
e
x
cos
θ
+
y
sin
θ
d
θ
=
2
π
I
0
(
x
2
+
y
2
)
{\displaystyle \int _{0}^{2\pi }e^{x\cos \theta +y\sin \theta }d\theta =2\pi I_{0}\left({\sqrt {x^{2}+y^{2}}}\right)}
∫
−
∞
∞
(
1
+
x
2
ν
)
−
ν
+
1
2
d
x
=
ν
π
Γ
(
ν
2
)
Γ
(
ν
+
1
2
)
{\displaystyle \int _{-\infty }^{\infty }\left(1+{\frac {x^{2}}{\nu }}\right)^{-{\frac {\nu +1}{2}}}\,dx={\frac {{\sqrt {\nu \pi }}\ \Gamma \left({\frac {\nu }{2}}\right)}{\Gamma \left({\frac {\nu +1}{2}}\right)}}}
(for ν > 0probability density function of Student's t -distribution )If the function f bounded variation on the interval [a ,b ] , then the method of exhaustion provides a formula for the integral:
∫
a
b
f
(
x
)
d
x
=
(
b
−
a
)
∑
n
=
1
∞
∑
m
=
1
2
n
−
1
(
−
1
)
m
+
1
2
−
n
f
(
a
+
m
(
b
−
a
)
2
−
n
)
.
{\displaystyle \int _{a}^{b}{f(x)\,dx}=(b-a)\sum \limits _{n=1}^{\infty }{\sum \limits _{m=1}^{2^{n}-1}{\left({-1}\right)^{m+1}}}2^{-n}f(a+m\left({b-a}\right)2^{-n}).}
The "sophomore's dream ":
∫
0
1
x
−
x
d
x
=
∑
n
=
1
∞
n
−
n
(
=
1.29128
59970
6266
…
)
∫
0
1
x
x
d
x
=
−
∑
n
=
1
∞
(
−
n
)
−
n
(
=
0.78343
05107
1213
…
)
{\displaystyle {\begin{aligned}\int _{0}^{1}x^{-x}\,dx&=\sum _{n=1}^{\infty }n^{-n}&&(=1.29128\,59970\,6266\dots )\\[6pt]\int _{0}^{1}x^{x}\,dx&=-\sum _{n=1}^{\infty }(-n)^{-n}&&(=0.78343\,05107\,1213\dots )\end{aligned}}}
Johann Bernoulli .
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![{\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\sqrt {\frac {\pi }{a}}}\exp \left[{\frac {b^{2}-4ac}{4a}}\right]}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/9aac3de5d20ec70a35c8d52bafaa6a82c148b566)
![{\displaystyle {\begin{aligned}\int _{0}^{1}x^{-x}\,dx&=\sum _{n=1}^{\infty }n^{-n}&&(=1.29128\,59970\,6266\dots )\\[6pt]\int _{0}^{1}x^{x}\,dx&=-\sum _{n=1}^{\infty }(-n)^{-n}&&(=0.78343\,05107\,1213\dots )\end{aligned}}}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/3eb1c1f34b4606f5ebf9e88ecdb80a7681b8fcce)