Question

Definite integral problem

Answer 1

Prove that:

\[\large{\int\limits_{0}^{\infty}(x^m*e^{-x}*\sin{x}~ dx) = \cfrac{m!}{2^{\cfrac{(m+2)}{2}}}\sin[(m+1)\cfrac{\pi}{4}]}\]

Answer 2

for m = 0,1,2,...

Answer 3

@ganeshie8 @Miracrown @mathmale @mathslover

I have some thought of this question

Answer 4

I am thinking of changing the integral to this:

\[\large{\int\limits_{0}^{\infty}(x^m*e^{-x}*\sin{x}~ dx) = \int\limits_{0}^{\infty}(x^m * Im(e^{(i-1)x})dx}\]

Answer 5

I am thinking of calculating this integral using integration by parts:

\[\large{\int\limits_{0}^{\infty}(x^m*e^{-x}*\sin{x}~ dx) = Im(\int\limits_{0}^{\infty}(x^m * (e^{(i-1)x})dx})\]

Answer 6

I did this:

\[\large{I~~\text{(say)} = \int\limits_{0}^{\infty}(x^m * e^{(i-1)x})dx}\]

Answer 7

\(\large \int_0^\infty f(t)dt =\lim_{x-\infty \int_0^x f(t)dt}\)

Answer 8

ohh ok @ikram002p , let me give it a try :

\[\large{\int\limits_{0}^{t}(x^m * e^{(i-1)x})dx}\]

\[\large{\int\limits_{0}^{t}(x^m~~\text{(I function)} * e^{(i-1)x}~~\text{(II function)})dx}\]

Answer 9

\[\large{x^m*(\int e^{(i-1)x}*dx) - \int(\int(e^{(i-1)x}*dx)*m*x^{m-1}*dx}\]

Answer 10

First I am going to calculate this:

\[\large{\int e^{(i-1)x}dx} = I_1\]

\[\large{I_1 = \cfrac{e^{(i-1)x}}{(i-1)}}\]

Answer 11

So,

\[\large{x^m*(\int e^{(i-1)x}*dx) - \int(\int(e^{(i-1)x}*dx)*m*x^{m-1}*dx}\]

\[\large{= x^m * \cfrac{e^{(i-1)x}}{(i-1)} - m\int(\cfrac{e^{(i-1)x}}{(i-1)} *x^{m-1}dx) }\]

\[\large{= x^m * \cfrac{e^{(i-1)x}}{(i-1)} - \cfrac{m}{i-1}* \int(e^{(i-1)x}*x^{m-1}dx)}\]


Should I apply integration by parts for second integral again @ikram002p

Answer 12

Let:

\[\large{I_2 = \int(e^{(i-1)x}*x^{m-1}*dx)}\]

\[\large{I_2 = \int(e^{(i-1)x}~~\text{(II function)}*x^{m-1}~~\text{(I function)}*dx)}\]

Answer 13

So,

\[\large{I_2 = x^{m-1}\int e^{(i-1)x}dx - \int (\int e^{(i-1)x} dx)*(m-1)*x^{m-2} dx}\]

I don't think this is going to help much.

Answer 14

I would get a similar integral as \(\large{I_1}\)

Answer 15

but it would be yet a lot different

Answer 16

Can gamma integral work ?

Answer 17

Because the limits and the terms to be integrated are very similar to it.

Answer 18

will there is direct epanding for
\(\int x^n e^x\)
u can directly use it

Answer 19

Ohh ok can you tell me the formula pls.

Answer 20

expanding *
sry for laging but im studing :D
ill write down

Answer 21

ohh okay take your time

Answer 22

\(\large \int x^n e^a dx =e^{ax}\sum_{n=0}^n (-1)^k \frac {n!}{(n-k)!} .\frac {x^{n-k}}{a^{ k+1 }} \)

Answer 23

Thanks @ikram002p I will try solving using this.

Answer 24

yeah it need work abit
s u would have this integral
\(\large \lim_{x-\infty}\frac{1}{2i}\int_0^x x^m e^{x(i-1) } dx -\lim_{x-\infty} \frac{1}{2i}\int_0^x x^m e^{-x(i+1) } dx\)

Answer 25

good luck !

Answer 26

OMG My goodness i haven't seen such big integrals ever in my whole life

Answer 27

Then you may need to see this haha @No.name
http://math.stackexchange.com/questions/815863/1-d-definite-integral

Answer 28

WOW , lOl

Answer 29

xD

Answer 30

Okay I tried it with your method but I think I did some mistake somewhere. So, I tried to solve it other way :

I am going to do this:

I assume that,

\[\large{I_n = \int \limits_{0}^{\infty} x^m e^{(i-1)x}dx }\]

\[\large{A_n = \int \limits_{0}^{\infty} x^m e^{-x} \cos x dx}\]

\[\large{B_n = \int \limits_{0}^{\infty} x^m e^{-x} \sin x dx}\]

Now, clearly,

\(\large{I_n = A_n + iB_n}\) with \(\large{n \ge 1}\)

First, I am going to solve \(\large{I_n}\) using by parts:

\(\huge{I_n = x^m\cfrac{e^{(i-1)x}}{i-1} - \int \limits m*x^{m-1}*\cfrac{e^{(i-1)x}}{i-1}dx}\)

I am trying to solve it after that. I would be back after dinner.

Answer 31

*

\[\large{I_n = x^m\cfrac{e^{(i-1)x}}{i-1} - \int \limits m*x^{m-1}*\cfrac{e^{(i-1)x}}{i-1}dx}\]

Answer 32

Wonderful !! I got it

Answer 33

ERRATA: In place of I_n, A_n, B_n there should be I_m, A_m, B_m.


Now, my solution after the above step:

\[\large{I_m = \cfrac{-m}{i-1} I_{m-1}}\]

\[\large{\implies I_m = \cfrac{-m}{i-1}*\cfrac{-(m-1)}{i-1}I_{m-2}}\]

\[\large{\implies I_m = \cfrac{(-1)^m*m!}{(i-1)^m}I_{0}}\]

Now, the value of \(\large{I_{0}}\) can be easily calculated and is,

\[\large{I_{0} = \cfrac{-1}{i-1}}\]

Thus,

\[\large{I_m = \cfrac{(-1)^m * m!}{(i-1)^m}*\cfrac{(-1)}{i-1}}\]

\[\large{\implies I_m = \cfrac{(-1)^{m+1} * m!}{(i-1)^{m+1}}}\]

\[\large{\implies I_m = \cfrac{(-1)^{m+1}*m!}{(i-1)^{m+1}}*\cfrac{(i+1)^{m+1}}{(i+1)^{m+1}}}\]

\[\large{\implies I_m = \cfrac{(-1)^{m+1}*m!*(i+1)^{m+1}}{(i^2-1)^{m+1}}}\]

\[\large{\implies I_m = \cfrac{(-1)^{m+1}*m!*(i+1)^{m+1}}{(-2)^{m+1}}}\]

\[\large{\implies I_m = \cfrac{m!*(i+1)^{m+1}}{2^{m+1}}}\]

Answer 34

Also,

\[\large{B_m = \cfrac{1}{2i}(I_m - \overline{I_m})}\]

\[\large{\implies B_m = \cfrac{1}{2i} * \cfrac{m!}{2^{(m+1)}}((i+1)^{m+1} - (1-i)^{m+1})}\]

\[\large{\implies B_m = \cfrac{m!}{2^{m+2}i}*[2^ {\cfrac{m+1}{2}}((\cos{(m+1)\cfrac{\pi}{4}}} + i\sin{(m+1)\cfrac{\pi}{4}}) - \]
\[\large{(\cos{(m+1)\cfrac{\pi}{4}} - i\sin{(m+1)\cfrac{\pi}{4}}))}\]

\(\large{\implies B_m = \cfrac{m!}{2^{\cfrac{m+1}{2}}}*[\sin ({(m+1)\cfrac{\pi}{4}})]}\) ;
\(\large{m \ge 0}\)

Answer 35

There we have it .. *sigh*

Answer 36

@ganeshie8 @ikram002p @mathmale @hartnn

Can you please check this proof ?

Answer 37

excellent work!
i don't see any error when i went through it, method is perfect...but seeing that you got the final answer correctly, i don't think there would be any error.

Answer 38

when you learn laplace, you might find a shorter way of doing same integral :)

Answer 39

Thanks @hartnn !! I would surely have a look at Laplace.