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  1. زيد اليافعي
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    #1
    Hello

    Integration is one of the most interesting topics in Mathematics. It has a wide area of applications in very different aspects of engineering and science. There are numerous ways to tackle integration problems. For elementary ones please refer to the following for basic introduction to integral calculus .

    For the most part, I don't want you to be afraid by the word ''advanced'' the most important factor is practice and practice as long as you have basic knowledge of elementary integral calculus you should not get afraid. Some exercises here will require basic knowledge of certain properties of special functions which I will introduce before tackling certain problems (I will not focus on the proof) .Furthermore , basic background of complex variables will be such a great help here. I will try to leave certain problems for the reader with a final answer to try in your leisure time .

    Finally , I will try to post an exercise every day so if you find any mistake don't hesitate to inform me . Furthermore , if you have any comments , face any problem understanding something or you want to show me your work just send me a pm or post it here and I will try to respond as soon as possible .

    [new] I created a pdf for my tutorials .
    Last edited by ZaidAlyafey; November 9th, 2015 at 06:50.

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    #2 Thread Author
    1. Differentiation under the integral sign :


    This is one of the most commonly used techniques to solve numerous numbers of questions I will be using this technique to solve many other exercises in the coming posts





    Assume that we have the following function of two variables :


    $ \displaystyle \int ^{b}_a \, f(x,y) \, dx$


    Then we can differentiate it with respect to y provided that f has partial continuous derivative on a chosen interval.


    $ \displaystyle F'(y) = \int^{b}_{a} f_{y}(x,y)\,dx$





    Now using this in many problems is not that clear you have to think a lot to get the required answer because many integral questions are just in one variable so you add the second variable and assume it is a function of two variables .


    Assume we want to solve the following integral :



    $ \displaystyle \int ^{1}_{0} \, \frac{x^2-1}{\ln (x) }$


    Now that seems very difficult to solve but using this technique we can solve it easily no matter how much power is x raised to . So the crux move is to decide ,where to put the second variable ! So the problem with the integral is that we have a logarithm in the denominator which makes the problem so difficult to tackle !

    Remember that we can get a natural logarithm if we differentiate exponential functions i.e $F(a) = 2^a \Rightarrow \,\, F'(a) = \ln(a) \cdot 2^a$


    Applying this to our problem


    $ \displaystyle F(a)=\int ^{1}_{0} \, \frac{x^a-1}{\ln (x) }$


    Differentiate with respect to $a$


    $ \displaystyle F'(b)=\frac{d}{d a}\int^{1}_{0} \frac{x^a-1}{\ln(x)}\,\,dx$


    We see that the integral is a function with two variables $f(x,a)$


    So we take the partial derivative with respect to a ..



    $ \displaystyle F'(a)=\int^{1}_{0} \frac{\partial }{\partial a}\left(\frac{x^a-1}{\ln(x)}\right) \,\,dx $


    $ \displaystyle F'(a)=\int^{1}_{0} x^a\,\,dx$


    $ \displaystyle F'(a)=\frac{x^{a+1}}{a+1} \bigl]^1_0$


    $ \displaystyle F'(a)=\frac{1}{a+1}$


    Now integrate wrt to a to get $ \displaystyle F(a)$


    $ \displaystyle F(a)=\ln{(a+1)}+C$


    $ \displaystyle F(0)=\ln{1}+C\,\,\text{hence : } C=0 $


    $ \displaystyle \int^{1}_{0} \frac{x^a-1}{\ln(x)}\,\,dx=\ln{(a+1)}$


    By this powerful rule we were not only able to solve the integral we also found a general formula for some $ \displaystyle a$ .


    Now to solve our original integral put $ \displaystyle a=2$


    $ \displaystyle \int^{1}_{0} \frac{x^2-1}{\ln(x)}\,\,dx=\ln{(2+1)}=\ln(3) $





    [HW]
    $ \displaystyle \int^{\infty}_{0}\frac{\sin (x) }{x}\,dx$


    To Be Continued ...
    Last edited by ZaidAlyafey; June 5th, 2013 at 21:47.

  3. زيد اليافعي
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    #3 Thread Author
    1.Differentiation under the integral sign (continued)


    It is not always easy to find the function of two variables in which to differentiate. So that requires insight and ability to foresee the function.




    Here is an example , find the following integral


    $ \displaystyle \int_0^{\frac{\pi}{2}} \frac{x}{\tan x}dx $





    So where do we put the variable $ \displaystyle a$ here , that doesn't seem to be straight forward , how do we proceed ?


    Well, take a deep breath , as I said it is not always easy to deduce such a function , it might not be so straight forward .


    Let us try the following


    $$ F(a)= \int_0^{\frac{\pi}{2}} \frac{\arctan(a\tan(x))}{\tan (x)}dx $$


    Now differentiate with respect to $ \displaystyle a$ :


    $ \displaystyle F'(a)= \int_0^{\frac{\pi}{2}} \frac{1}{1+(a\tan(x) )^2}dx $


    We will show later that


    $ \displaystyle \int_0^{\frac{\pi}{2}} \frac{1}{1+(a\tan(x) )^2}dx= \frac{\pi}{2(1+a)}$


    $ \displaystyle F'(a) = \frac{\pi}{2(1+a)}$


    Now Integrate both sides


    $ \displaystyle F(a)= \frac{\pi}{2}\ln(1+a) +C $


    Now we need to substitute $a=0$ in order to find $C = 0$



    $ \displaystyle \int_0^{\frac{\pi}{2}} \frac{\arctan(a\tan(x))}{\tan (x)}\,dx = \frac{\pi}{2}\ln(1+a) $



    Put $a =1$ in order to get our original integral



    $ \displaystyle \int_0^{\frac{\pi}{2}} \frac{x}{\tan (x)}\,dx = \frac{\pi}{2}\ln(2) $
    Last edited by ZaidAlyafey; June 22nd, 2013 at 21:52.

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    #4 Thread Author
    Differentiation under the integral sign (continued):

    I already gave this as a practice problem I will solve it now to check your solution :

    We are given the following :


    $$ \int^{\infty}_0 \frac{\sin(x) }{x}\, dx $$


    This problem can be solved by many ways , but here we will try to solve it by differentiation .

    So as I described earlier in the previous examples it is generally not so easy to find the function with two variables .

    Actually this step might require trial and error techniques until we get the desired result , so don't just give up if an approach merely doesn't work !.

    $ \displaystyle F(a)=\int^{\infty}_0 \frac{\sin(ax) }{x}\, dx$


    Let us try this one :


    If we differentiated with respect to a we get the following :


    $ \displaystyle F'(a)=\int^{\infty}_0 \cos(ax) \, dx$


    But unfortunately this integral doesn't converge , so this is not the correct one .

    Well, that seemed hopeless , but you should benefit from mistakes . The previous integral will converge if there is an exponential (This is merely the Laplace transform which I will illustrate later ... ).


    So let us try the following :


    $ \displaystyle F(a)=\int^{\infty}_0 \frac{\sin(x) e^{-ax}}{x}\, dx$


    Take the derivative to get :


    $ \displaystyle F'(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dx$


    Now this is easy to solve we can use integration by parts twice to get the following :


    $ \displaystyle F'(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dx=\frac{-1}{a^2+1}$


    Now integrate both sides :


    $ \displaystyle F(a)=-\arctan(a)+C$


    Now to find the value of the constant will take the limit as a grows very large :



    $ \displaystyle C=\lim_{a\to \infty }F(a) +\arctan(a)= \frac{\pi}{2}$


    So we get our F(a) as the following:


    $ \displaystyle F(a)=-\arctan(a)+ \frac{\pi}{2}$


    Now for the particular value of a = 0 we have :


    $ \displaystyle \int^{\infty}_0 \frac{\sin(x) }{x}=\frac{\pi}{2}$


    Please if anything not clear you can send me a pm ...

    In the next post I will start Hyperbolic Integration.

  5. زيد اليافعي
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    #5 Thread Author
    2.Hyperbolic Integration :

    Hyperbolic functions are very interesting , they are used to solve several integration problems. The most interesting thing about them is their relation to geometric functions through the see here.

    It is really power full that you can switch between trigonometric functions and hyperbolic functions and vice versa , this requires a basic knowledge of complex numbers.

    We know from the general definition of hyperbolic functions that :

    $$\sinh(x) = \frac{e^x-e^{-x}}{2}$$

    $$\cosh(x) = \frac{e^x+e^{-x}}{2}$$

    Well, that seems interesting as it seems very close to the Euler's formula for sine and cosine :


    $$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$$

    $$\cos(x) = \frac{e^{ix}+e^{-ix}}{2}$$

    It can be deduced easily that we can use the following to convert from trigonometric to hyperbolic functions :

    $$\cos(ix)= \cosh(x)$$

    $$\sin(ix)= i\sinh(x)$$

    Woo , that is really fabulous so we can apply this to several problems to get rid of the complex i which is really annoying .

    Here is an interesting problem :

    $ \displaystyle \int - \left( \frac{\pi \csc\left(\pi (\frac{-1-ix}{2})\right)}{4}+\frac{\pi \csc\left(\pi (\frac{-1+ix}{2})\right)}{4}\right)\, dx$

    That seems very very very .... complicated , but wait don't be cheated by the complex view of a problem . Think about our hyperbolic friend.

    Here we can apply basic trigonometric simplifications :

    $ \displaystyle \int \frac{-\pi}{4}\left(\frac{1}{\sin\left(\frac{-\pi}{2})\cos(\frac{i\pi x}{2}\right)}+\frac{1}{\sin\left(\frac{-\pi}{2}\right)\cos\left(\frac{i\pi x}{2}\right)} \right) \, dx$

    Now we see how to apply our magic trick

    $ \displaystyle \int \frac{\pi}{4}\left(\frac{1}{\cosh\left(\frac{\pi x}{2}\right)}+\frac{1}{\cosh\left(\frac{\pi x}{2}\right)} \right) \, dx$

    This can be further simplified to get the following :

    $ \displaystyle \frac{\pi}{2}\int \text{sech}\left(\frac{\pi x}{2}\right)\,dx=2\arctan \left( \tanh\left(\frac{\pi x}{4}\right) \right)+C $

    Ooh , wait , hang on How is that ?!!

    Well, differentiate the right side to see what is going on :

    $ \displaystyle 2\frac{\frac{\pi}{4} \text{sech}^2 \left(\frac{\pi x}{4}\right)}{1+\tanh^2 \left(\frac{\pi x}{4}\right)}$


    $ \displaystyle \frac{\pi}{2}\left(\frac{1}{\cosh^2\left(\frac{\pi x}{4}\right) +\sinh^2\left(\frac{\pi x}{4}\right)} \right)$


    Further simplification will do the task :

    $ \displaystyle \frac{\pi}{2}\left(\frac{1}{\cosh^2(\frac{\pi x}{4}) +\cosh^2(\frac{\pi x}{4})-1}\right)=\frac{\pi}{2}\left(\frac{1}{\cosh(\frac{\pi x}{2})}\right)= \frac{\pi}{2}\text{sech}\left(\frac{\pi x}{2}\right)$


    To be continued ...

  6. زيد اليافعي
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    #6 Thread Author
    Hyperbolic Integration (continued)

    As we have seen in the previous post hyperbolic functions can be so helpful as they help us simplify a lot of operations in a way that is really neat. Also, they help us handle complex variables to get a result that is completely free of complex numbers. This is basically because of the ability to convert from trigonometric functions though Euler formula.

    Working with hyperbolic functions for the most part requires a lot of practice. It might not be so clear that the function can be integrated , this will help us in complex integration in the future .

    Let us have the following example :

    prove that :


    $$I=\int -\frac{1}{2} \left( \csc(-\alpha-ix)+\csc(-\alpha+ix)\right)= \arctan\left(\csc(\alpha) \sinh(x) \right) +C$$


    $ \displaystyle \frac{1}{2}\int \csc(\alpha+ix)+\csc(\alpha-ix) \,dx$


    $ \displaystyle \frac{1}{2}\int \frac{1}{\sin(\alpha+ix)}+\frac{1}{\sin(\alpha-ix)}\, dx$


    Applying basic geometry identities :


    $ \displaystyle =\frac{1}{2}\int \left(\frac{1}{\sin(\alpha)\cosh(x) +i\sinh(x) \cos(\alpha)}+\frac{1}{\sin(\alpha)\cosh(x) -i\sinh(x) \cos(\alpha)}\right)\, dx$


    Notice that the denominators are the complex conjugate of each other :


    $ \displaystyle \frac{1}{2}\int \left(\frac{2\sin(\alpha)\cosh(x)}{\sin^2(\alpha) \cosh ^2(x) +\sinh^2(x) \cos^2(\alpha)}\right)\, dx=\, \int \left(\frac{\sin(\alpha)\cosh(x)}{\sin^2(\alpha)(1+\sinh^2(x)) +\sinh^2(x) (1-\sin^2(\alpha))}\right)\, dx$


    $ \displaystyle \int \left(\frac{\sin(\alpha)\cosh(x)}{\sin^2(\alpha) +\sinh^2(x)}\right)\, dx= \int \frac{\csc(\alpha)\cosh(x)}{ 1+\csc^2(\alpha)\sinh^2(x)}\, dx$


    $ \displaystyle I= \arctan\left(\csc(\alpha) \sinh(x) \right) +C$


    I will start Laplace transform in the next post ...
    Last edited by ZaidAlyafey; February 2nd, 2013 at 23:47. Reason: simplify readibility

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    #7 Thread Author
    [U][B]3.Integration by Laplace Transform (here.


    Let us see some examples on integration:


    Find the following integral :

    $ \displaystyle \int^{\infty}_{0}e^{-2t}t^3\,dt $

    We can directly use the above formula

    $ \displaystyle \int ^{\infty}_0 e^{-st}t^n \, dt=\frac{n!}{s^{n+1}}$

    here we have s= 2 , and n =3 , so directly we get :

    $ \displaystyle \int^{\infty}_{0}e^{-2t}t^3\,dt= \frac{3!}{2^{3+1}}=\frac{3}{8} $

    So , we see that is becoming easy to find ...


    Now let us think about the Laplace Inverse :


    So, basically you are given F(s) and we want to get f(t) this is denoted by :

    $$\mathcal{L}^{-1}(F(s))= f(t)\text{ or }\mathcal{L}^{-1}(\mathcal{L}(f(t)))= f(t)$$

    We are given a function in the variable s and we want to transform it into another in the variable t .


    Suppose we have the following examples :

    Find the Laplace inverse of the following :

    1) $ \displaystyle \frac{1}{s^3}$

    2) $ \displaystyle \frac{s}{s^2+4}$

    1- we can use the result in (2) so we have

    $ \displaystyle \mathcal{L}(t^2)= \frac{2!}{t^3}\,\, \Rightarrow \,\, \frac{1}{2}\mathcal{L}(t^2)= \frac{1}{s^3}$

    Now take the inverse to both sides :

    $ \displaystyle \frac{t^2}{2}= \mathcal{L}^{-1}\left(\frac{1}{s^3}\right)$

    2- we can use the result in (3) so we have

    $ \displaystyle \cos(2t)=\mathcal{L}^{-1}\left(\frac{s}{s^2+4}\right)$


    [HW.1] Find the Laplace of $\sin(at)$

    [HW.2] Find the inverse Laplace of $ \displaystyle \frac{1}{s^{n+1}}$

    To be continued ...
    Last edited by ZaidAlyafey; February 3rd, 2013 at 06:35.

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    #8 Thread Author
    Integration by Laplace Transform (continued)

    3.2.1.Interesting results :


    Prove the following :

    $$B(x+1,y+1)=\int^{1}_{0}t^{x}\, (1-t)^{y}\,dt= \frac{\Gamma(x+1)\Gamma{(y+1)}}{\Gamma{(x+y+2)}}$$ B is and $\Gamma $ is

    Note : Just as a basic introduction to special functions, you can think of gamma function as the following : $n!=\Gamma{(n+1)}$ we can extend n to exist in the whole complex plane {except n is a negative integer }.

    The beta function is so interesting , we will explain some of its basic properties later .It can be used to solve many integrals.

    For the proof :


    We have the defnition (4) in the previous post about convolution .

    Let us choose some functions f , and g :


    $f(t) = t^{x} \,\, , \, g(t) = t^y$

    By substituting in (4) we get the following :


    $ \displaystyle (t^x*t^y)= \int^{t}_0 s^{x}(t-s)^{y}\,ds \text{ ---- (*)}$


    We have from definition (5) that :


    $\mathcal{L}\left(t^x*t^y\right)= \mathcal{L}(t^x) \mathcal{L}(t^y ) $


    Now we can use the result (2) to deduce



    $ \displaystyle \mathcal{L}\left(t^x*t^y\right)= \frac{x!\cdot y!}{s^{x+y+2}}$


    Notice that we need to find the inverse of Laplace $\mathcal{L}^{-1}$


    $ \displaystyle \mathcal{L}^{-1}\left(\mathcal{L}(t^x*t^y)\right)=\mathcal{L}^{-1}\left( \frac{x!\cdot y!}{s^{x+y+2}}\right)=t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!}$


    So we have the following :


    $ \displaystyle (t^x*t^y) =t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!}$


    Now substitute in --(*) we get :


    $ \displaystyle t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!} = \int^{t}_0 s^{x}(t-s)^{y}\,ds$


    This looks good , put t=1 we get :


    $ \displaystyle \frac{x!\cdot y!}{(x+y+1)!} = \int^{1}_0 s^{x}(1-s)^{y}\,ds$


    By using that $n! = \Gamma{(n+1)}$


    we arrive happily to our formula :


    $$ \int^{1}_0 s^{x}(1-s)^{y}\,ds= \frac{\Gamma(x+1)\Gamma{(y+1)}}{\Gamma{(x+y+2)}}$$

    To be continued ...

  9. زيد اليافعي
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    #9 Thread Author
    Integration by Laplace Transform (continued)

    3.2.2.Interesting results :



    Prove the following :

    $$\int^{\infty}_0 \frac{f(t)}{t}\,dt=\, \int^{\infty}_0 \, \mathcal{L}(f(t))\, ds\text{ ----(6)}$$

    This is a very powerful rule it is basically saying that if we have a function divided by its independent variable and we integrated it in the half positive plane then we can transform it into an integral wrt to the Laplace transform of the function .


    So Let us prove it :


    we will start from the right hand side of (6)

    we know from the definition --(1) that :


    $ \displaystyle \int^{\infty}_0 \, \mathcal{L}(f(t))\, ds= \int^{\infty}_0 \,\left( \int^{\infty}_0e^{-st}\, f(t) \,dt\right)\, ds$


    Now by the we can rearrange* the double integral :


    $ \displaystyle \int^{\infty}_0 \, f(t) \left(\int^{\infty}_0 e^{-st}\,ds\right)\,dt \text{ ----(**)}$


    The integral inside the parenthesis :


    $ \displaystyle \int^{\infty}_0 e^{-st}\,ds $


    I know this is elementary ,but you may realize this is the Laplace transform but now the function is f(s) so basically we are just evaluating F(t) when f(s) =1 .


    $ \displaystyle \int^{\infty}_0 e^{-st}\,ds = \mathcal{L}(1)= \frac{1}{t}$


    Now substitute this value in the integral in (**)


    $ \displaystyle \int^{\infty}_0 \, \frac{f(t)}{t}\,dt $


    which is the left hand side of ----(6) as just required ...

    This is just a fabulous rule , it can be used to simplify many computations ...


    Let us find the following integral :


    $$\int^{\infty}_0 \frac{\sin(t)}{t}\,dt$$


    This is not the first time we see this integral and not the last . We have seen that we can find it using differentiation under the integral sign .

    But you are just about to see the power of definition (6) [Doe that look like a movie trailer ] , never mind .

    Back to our integral , so the integral is of the type $\frac{f(t)}{t}$ so we can easily find it .


    $ \displaystyle \int^{\infty}_0 \frac{sin(t)}{t}\,dt= \int^{\infty}_{0}\mathcal{L}(\sin(t))\,ds$


    Now what is the Laplace transform of $\sin(t) $ , did you do your homework ?

    $$\mathcal{L}(\sin(at)) = \frac{a^2}{s^2+a^2}$$

    So putting a =1 we get :

    $ \displaystyle \mathcal{L}(\sin(t)) = \frac{1}{s^2+1}$

    Just substitute in our integral :


    $ \displaystyle \int^{\infty}_0 \frac{ds}{1+s^2}= \tan^{-1}(s)|_{s=\infty}-\tan^{-1}(s)|_{s=0}=\frac{\pi}{2}$


    Just as expected ...


    In the next post , I will explain the Gamma function .

    * We must prove that we can rearrange the double integral , I didn't go into the details.

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    #10 Thread Author
    4.Integration using special functions

    4.1.Gamma Function :


    is really interesting it is used to solve many interesting integrals, here we try to define some basic properties , prove some of them and take some examples.

    Definition :


    $$ \Gamma{(x+1)} = \int^{\infty}_0 e^{-t}\, t^{x}\, dt \text{ ---(1)} $$

    For the first glance that just looks like the Laplace Transform , actually they are closely related .

    So let us for simplicity assume that x=n where $ \displaystyle n\geq 0 \text{ , n }$ (is an integer )and substitute in (1) we get :

    $ \displaystyle \Gamma{(n+1)} = \int^{\infty}_0 e^{-t}\, t^{n}\, dt $

    Well, we can use the Laplace transform - you might revise if you forgot -

    $ \displaystyle \int^{\infty}_0 e^{-t}\, t^{n}= \frac{n!}{s^n+1}|_{s=1}= n!$

    So we see that there is a relation between the gamma function and the factorial ..

    We will assume for the time being that the gamma function is defined as the following

    $ \displaystyle n!= \Gamma{(n+1)} $.

    By this definition n\geq 0 where n is any positive integer which is pretty limited but surely this definition will be soon replaced by a stronger one.

    Let us have some Examples :

    Find the following integrals :

    $ \displaystyle \int^{\infty}_0 \, e^{-t}t^4\, dt $

    By definition (1) this can be replaced by

    $ \displaystyle \int^{\infty}_0 \, e^{-t}t^4\, dt =\Gamma(4+1)= 4! = 24 $

    Pretty good let us continue with some more examples :

    1) $ \displaystyle \int^{\infty}_0 \, e^{-t^2}t \, dt $

    2) $ \displaystyle \int^{1}_{0}\ln (t) \, t^2 \, dt $

    1) For the first one clearly we need a substitution before we go ahead :

    so let us start by putting $ \displaystyle x=t^2$ so the integral becomes :

    $ \displaystyle \frac{1}{2}\int^{\infty}_0 \, e^{-x}x^{\frac{1}{2}}\cdot x^{-\frac{1}{2}}\, dt =\frac{1}{2}\Gamma(1+0)=\frac{1}{2}$

    2) For the second we use the substitution $ \displaystyle t=e^{-\frac{x}{2}}$

    $ \displaystyle \int^{1}_{0}\ln (t) \, t^2 \, dt = -\frac{1}{4}\int^{\infty}_{0}e^{-\frac{3x}{2}}\cdot x\,dx$

    Using another substitution $ \displaystyle t=\frac{3x}{2}$ (Back to my favourite symbol )

    $ \displaystyle \frac{-1}{9}\int^{\infty}_0 e^{-x}\, x\, dx =\frac{- \Gamma(2)}{9}= \frac{-1}{9}$

    It is an important thing to get used to the symbol $\Gamma$. I am sure that you are saying (boring...) that this seems elementary , but my main aim here is to let you practice the new symbol and get used to solving some problems using it .

    [HW] prove $ \displaystyle \frac{\Gamma(5)\cdot \Gamma(2)}{\Gamma(7)}=\frac{1}{30}$

    [HW] Find the integral : $ \displaystyle \int^{\infty}_{0}e^{-\frac{1}{60}t}\, t^{20}\, dt $ in terms of the gamma function .



    In the next post I should extend the idea of a factorial ... you know this is only get exciting ...

    To be continued.




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