Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

Get extra R 320 off
USE CODE: CART20

				   plz sir   provied me full information about    gamma function.beta function, walli's function  .how i solve question from these type of function


6 years ago

Share

										Dear nitish
Gamma function
In mathematics, the Gamma function (represented by the capital Greek letter Γ) is an extension of the factorial function to realcomplex numbers. For a complex number z with positive real part the Gamma function is defined by and
$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt\;$
This definition can be extended by analytic continuation to the rest of the complex plane, except the non-positive integers.
If n is a positive integer, then
Γ(n) = (n − 1)!
showing the connection to the factorial function. Thus, the Gamma function extends the factorial function to the real and complex values of n.
Beta function

In mathematics, the beta function, also called the Euler integral of the first kind, is a special function defined by
$\mathrm{\Beta}(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt \!$
Properties

The beta function is symmetric, meaning that
$\Beta(x,y) = \Beta(y,x). \!$
It has many other forms, including:
$\Beta(x,y)=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)} \!$ $\Beta(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,d\theta, \qquad \textrm{Re}(x)>0,\ \textrm{Re}(y)>0 \!$ $\Beta(x,y) = \int_0^\infty\dfrac{t^{x-1}}{(1+t)^{x+y}}\,dt, \qquad \textrm{Re}(x)>0,\ \textrm{Re}(y)>0 \!$ $\Beta(x,y) = \sum_{n=0}^\infty \dfrac{{n-y \choose n}} {x+n}, \!$ $\Beta(x,y) = \frac{x+y}{x y} \prod_{n=1}^\infty \left( 1+ \dfrac{x y}{n (x+y+n)}\right)^{-1}, \!$ $\Beta(x,y) \cdot \Beta(x+y,1-y) = \dfrac{\pi}{x \sin(\pi y)}, \!$

Wallis Formula

The Wallis formula follows from the infinite product representation of the sine

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards,Askiitians ExpertsBadiuddin


6 years ago

# Other Related Questions on Integral Calculus

How many distinct values of 𝑥 satisfy the equation sin(𝑥) = 𝑥/2, where 𝑥 is in radians? (A) 1 (B) 2 (C) 3 (D) 4 or more

plot the graph of sin(x) and x/2. From just observing it we can see that it intersects at 3 points and hence the answer is 3.

 Riddhish Bhalodia 7 months ago

they give sin(x)=x/2 0 is the one solution because sin(0)=0 so one solution exist for this question if you take pi/2 or pi/4 you have to p[rove that sin pi/2=pi/4 or sin pi/8=pi/16 which is ...

 SREEKANTH 4 months ago
lt 1/n[tanpie/4n+tan2pie/4n+.....+tan npie/4n] n-infinity

This questions is from integration as a limit of sum This quantity is equivalent to Thanks

 Nishant Vora 2 months ago

write x^3×cosx^3dx as: x*x^2*cosx^3dx => let x^2 = u 2xdx = du I = u*cos(u^(3/2)) Apply now by – part easily then after substitute u in the end.

 Vikas TU 3 months ago
the number of normals to a parabola y^2 = 8x through (2,1) is

Equation of normal y = mx – 2am – am 3 here a= 2 from parabola ​Passes through (2,1) 1= 2m – ​4m – ​2m ​3 2m ​3 ​+ 2m + 1 =0 Derivate of above function is 6m ​2 + 2...

 Harsh Patodia one month ago
Lim x->2 (2/(x+2) + 1/x2-2x+4 - 24/(x3+8) this is the question

put x = 2 in Lim x->2 (2/(x+2) + 1/x2-2x+4 - 24/(x3+8) =>2/4 + ¼ – 24/16 => 0.5 + 0.25 – 1.5 = > -.75 or -3/4

 Vikas TU 3 months ago
what is formula for orthocentre of a triangle or how to calculate it?

Direct formula is there but it is not useful as it is in terms of tanA , tanB , tanC which is not useful Easiest way to find orthocentre: i) It is the point of concurrency of the 3 altitudes...

 Harsh Patodia one month ago
View all Questions »

• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: R 15,000
• View Details
Get extra R 3,000 off
USE CODE: CART20

Get extra R 320 off
USE CODE: CART20

More Questions On Integral Calculus