Evaluate this definite integral

In summary, the integral $$ \displaystyle \int_0^{\infty} e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2} da $$ can be simplified by using the substitution $a = w$, which reduces it to finding the integral $$\displaystyle \int_{-\infty}^{\infty} \frac{w \sin(b w) - \cos(bw)}{1+w^2} dw$$ where $b > 0$. By taking the limit as $r \to 0+$, we can show that this integral is equal to $0$, thus the original integral also does not exist.
  • #1
utkarshakash
Gold Member
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Homework Statement


$$ \displaystyle \int_0^{\infty} e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2} da $$

Homework Equations



The Attempt at a Solution



Evaluating this using integration by parts will be a cumbersome process and I don't even think that would give me the answer. Substitutions aren't going to make this problem easier. What other methods should I apply?
 
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  • #2
simplify or change the form of the integrand ... i.e. look for an identity involving the trig functions.
the euler formula perhaps ... what have you tried?

[edit] Ugh - looking up the indefinite integrals ... it's nasty.
See "trigonometric integrals"
http://en.wikipedia.org/wiki/Trigonometric_integral

Where does this come from?
 
Last edited:
  • #3
Is the variable of integration 'a' instead of 'x'?
 
  • #4
SteamKing said:
Is the variable of integration 'a' instead of 'x'?

Yes
 
  • #5
utkarshakash said:

Homework Statement


$$ \displaystyle \int_0^{\infty} e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2} da $$

Homework Equations



The Attempt at a Solution



Evaluating this using integration by parts will be a cumbersome process and I don't even think that would give me the answer. Substitutions aren't going to make this problem easier. What other methods should I apply?

Differentiating under the integral sign might help here, I think.

$$\displaystyle \frac{d}{dx}\int_0^{\infty} e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2} da = \int_0^{\infty} \frac{\partial}{\partial x} (e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2}) da = \int_0^{\infty}e^{-x}\cos ax da $$ which is non convergent.

So the original integral also does not exist.

I'm not 100% sure about this, though.
 
Last edited:
  • #6
utkarshakash said:
Yes
Then I don't understand the significance of the exponential term. Why is that not just a factor that can be taken outside the integral? Or is everything inside the integral supposed to be forming the exponent?
 
  • #7
utkarshakash said:

Homework Statement


$$ \displaystyle \int_0^{\infty} e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2} da $$

Homework Equations



The Attempt at a Solution



Evaluating this using integration by parts will be a cumbersome process and I don't even think that would give me the answer. Substitutions aren't going to make this problem easier. What other methods should I apply?

I looked at your profile, and so cannot believe this is a homework problem you have been given in a course. For that reason, I am going to chance giving you a complete solution, a practice I normally avoid.

Let's change your unfortunate and rather horrible notation. What you call 'x' I will call 'b', and what you call 'a' I will call 'w'. So, you want to perform the integration
[tex] \int_0^{\infty} e^{-b} \frac{w \sin(b w) - \cos(bw)}{1+w^2} \, dw [/tex]
First: the constant ##e^{-b}## plays no role, so omit it; you can just do the integral without it, then put it back again in the final answer.
Second: the integrand is an even function of ##w##, so ##\int_0^{\infty} = \frac{1}{2} \int_{-\infty}^{\infty}##, a fact that proves convenient later.

OK, so we want the integral
[tex] \int_{-\infty}^{\infty} \frac{ w \sin(bw) - \cos(bw)}{1+w^2} \, dw [/tex]
WLOG, assume b > 0. We have ##w \sin(bw) = w \text{Re}[-i e^{ibw}]## and ##-\cos(bw) = \text{Re}[-e^{ibw}]##, so the integrand is the real part of
[tex] F = \frac{-(1+iw) e^{ibw}}{1+w^2} = \frac{-(1+iw) e^{ibw}}{(1+iw)(1-iw)} =-i \frac{e^{ibw} }{w+i}. [/tex]
Thus, it is enough to compute
[tex] J = \int_{-\infty}^{\infty} \frac{e^{ibw}}{w+i} \, dw [/tex]This is an improper integral, so let's replace it by
[tex] K_r = \int_{-\infty}^{\infty} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw [/tex]
and then take the limit as ## r \to 0+##.

The integrand of ##K_r## has a simple pole in the complex w-plane at the point ##w = -i##. For ##b > 0## we may complete the contour in the upper half w-plane, to get
[tex] K_r = \lim_{R \to \infty} \int_{-R}^{R} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw
= \lim_{R \to \infty} \oint_{\Gamma} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw,[/tex]
where ##\Gamma## is the contour that goes along the real w-axis from ##w = -R+i0## to ##w = +R+i0##, then goes counterclockwise from ##R + i0## to ##-R + i0## along the semicircle of radius ##R## in upper half of the complex ##w##-plane. But the contour integral = 0 because the integrand is analytic inside ##\Gamma##. Thus, ##K_r = \lim_{R \to \infty} 0 = 0##, and this holds for all ##r > 0##. Taking the limit we have
[tex] \int_{-\infty}^{\infty} \frac{e^{ibw}}{w+i} \, dw = \lim_{r \to 0} K_r = \lim_{r \to 0} 0 = 0.[/tex]

Thus, we obtain
[tex] \int_{-\infty}^{\infty} \frac{ w \sin(bw) - \cos(bw)}{w^2+1} \, dw = 0,[/tex]
provided that we regard it as the limit where ##r \to 0+##.
 
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  • #8
Ray Vickson said:
I looked at your profile, and so cannot believe this is a homework problem you have been given in a course. For that reason, I am going to chance giving you a complete solution, a practice I normally avoid.

Let's change your unfortunate and rather horrible notation. What you call 'x' I will call 'b', and what you call 'a' I will call 'w'. So, you want to perform the integration
[tex] \int_0^{\infty} e^{-b} \frac{w \sin(b w) - \cos(bw)}{1+w^2} \, dw [/tex]
First: the constant ##e^{-b}## plays no role, so omit it; you can just do the integral without it, then put it back again in the final answer.
Second: the integrand is an even function of ##w##, so ##\int_0^{\infty} = \frac{1}{2} \int_{-\infty}^{\infty}##, a fact that proves convenient later.

OK, so we want the integral
[tex] \int_{-\infty}^{\infty} \frac{ w \sin(bw) - \cos(bw)}{1+w^2} \, dw [/tex]
WLOG, assume b > 0. We have ##w \sin(bw) = w \text{Re}[-i e^{ibw}]## and ##-\cos(bw) = \text{Re}[-e^{ibw}]##, so the integrand is the real part of
[tex] F = \frac{-(1+iw) e^{ibw}}{1+w^2} = \frac{-(1+iw) e^{ibw}}{(1+iw)(1-iw)} =-i \frac{e^{ibw} }{w+i}. [/tex]
Thus, it is enough to compute
[tex] J = \int_{-\infty}^{\infty} \frac{e^{ibw}}{w+i} \, dw [/tex]This is an improper integral, so let's replace it by
[tex] K_r = \int_{-\infty}^{\infty} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw [/tex]
and then take the limit as ## r \to 0+##.

The integrand of ##K_r## has a simple pole in the complex w-plane at the point ##w = -i##. For ##b > 0## we may complete the contour in the upper half w-plane, to get
[tex] K_r = \lim_{R \to \infty} \int_{-R}^{R} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw
= \lim_{R \to \infty} \oint_{\Gamma} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw,[/tex]
where ##\Gamma## is the contour that goes along the real w-axis from ##w = -R+i0## to ##w = +R+i0##, then goes counterclockwise from ##R + i0## to ##-R + i0## along the semicircle of radius ##R## in upper half of the complex ##w##-plane. But the contour integral = 0 because the integrand is analytic inside ##\Gamma##. Thus, ##K_r = \lim_{R \to \infty} 0 = 0##, and this holds for all ##r > 0##. Taking the limit we have
[tex] \int_{-\infty}^{\infty} \frac{e^{ibw}}{w+i} \, dw = \lim_{r \to 0} K_r = \lim_{r \to 0} 0 = 0.[/tex]

Thus, we obtain
[tex] \int_{-\infty}^{\infty} \frac{ w \sin(bw) - \cos(bw)}{w^2+1} \, dw = 0,[/tex]
provided that we regard it as the limit where ##r \to 0+##.

I appreciate your effort and patience for posting the complete solution. However, your method was too complex for me. I don't know anything about contours and related terms. The integral which I posted here is a part of another integral which forms the original problem and here's it.

Calculate [itex]\displaystyle \int_0^{\infty} e^{-x} \dfrac{\sin ax}{x} dx [/itex]

Using Leibnitz's theorem for differentiation, I get
$$\displaystyle \int_0^{\infty} e^{-x} \cos ax dx $$
For evaluating this integral, I used the property(which is provided as a hint to the solution in my textbook) that
[itex]\displaystyle \int e^{ax} \cos bx dx = \dfrac{e^{ax}(a\cos bx + b \sin bx)}{a^2+b^2} [/itex]

Substituting appropriate values of a and b in the given formula, I get the integral posted in the thread.
 
  • #9
You need to plug the limits in for ##x##. That'll simplify the integrand to the point where it's easy to integrate with respect to ##a##.
 
  • #10
Curious3141 said:
Differentiating under the integral sign might help here, I think.

$$\displaystyle \frac{d}{dx}\int_0^{\infty} e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2} da = \int_0^{\infty} \frac{\partial}{\partial x} (e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2}) da = \int_0^{\infty}e^{-x}\cos ax da $$ which is non convergent.

So the original integral also does not exist.

I'm not 100% sure about this, though.

This reasoning is not correct. If you consider the integral ##\displaystyle \int_0^{\infty} \frac{\sin (ax)}{x}\,dx## and differentiate with respect to ##a##, you get a non convergent integral but the original integral does exist!

BTW, @utkarshakash, how did you get to the integral in OP from the original integral?
 
  • #11
Pranav-Arora said:
This reasoning is not correct. If you consider the integral ##\displaystyle \int_0^{\infty} \frac{\sin (ax)}{x}\,dx## and differentiate with respect to ##a##, you get a non convergent integral but the original integral does exist!

BTW, @utkarshakash, how did you get to the integral in OP from the original integral?

Thank you. Yes, that's why I stated I was unsure if it was correct in my first post, but I posted it in case someone else could use the idea to come up with a correct solution.
 
  • #12
utkarshakash said:
Using Leibnitz's theorem for differentiation, I get
$$\displaystyle \int_0^{\infty} e^{-x} \cos ax dx $$
.
From there, can you not express the cos in complex exponentials and just integrate?
 
  • #13
  • #14
haruspex said:
From there, can you not express the cos in complex exponentials and just integrate?
Or integrate by parts. However, the OP doesn't even need to do that as the hint provided gives the result of the indefinite integration. The problem is that by not plugging in the limits for ##x##, one is left with a complicated integrand in the last integration needed to get to the final solution.
 

1. What is a definite integral?

A definite integral is a mathematical concept used to calculate the area under a curve in a specific interval on a graph. It is represented by the symbol ∫ and has a lower and upper limit.

2. How is a definite integral evaluated?

A definite integral is evaluated by using either the fundamental theorem of calculus or integration techniques such as substitution, integration by parts, or partial fraction decomposition.

3. What is the difference between a definite and indefinite integral?

A definite integral has specific limits of integration and gives a numerical value, while an indefinite integral has no limits and gives a function as a result.

4. Can a definite integral have a negative value?

Yes, a definite integral can have a negative value if the function being integrated has negative values within the given interval. This means that the area under the curve in that interval is below the x-axis.

5. What are some real-life applications of definite integrals?

Definite integrals are used in various fields, such as physics, engineering, economics, and statistics, to calculate quantities such as displacement, work, profit, and probability.

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