Integral e^x.(lnx)dx=?

  • Thread starter sumerman
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In summary, the anti-derivative of your function involves the exponential integral function, which is defined here:
  • #1
sumerman
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i have tried some methods like uv - integral vdu but can't reach the answer
 
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  • #2
You won't be able to do it.
 
  • #4
Try to take the derivative of this with respect to $x$, and see what do you get:

[tex]e^x\left[\ln x-\sum_{i=1}^{\infty}(i-1)!x^{-i}\right][/tex]
 
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  • #5
cute.
 
  • #6
tony.c.tan said:
Try to take the derivative of this with respect to $x$, and see what do you get:

[tex]e^x\left[\ln x-\sum_{i=1}^{\infty}(i-1)!x^{-i}\right][/tex]

Supercool!

Would you show what techniques are useful to get that anti-derivative?
 
  • #7
tony.c.tan said:
Try to take the derivative of this with respect to $x$, and see what do you get:

[tex]e^x\left[\ln x-\sum_{i=1}^{\infty}(i-1)!x^{-i}\right][/tex]

Woh! a series that diverges for every x ... what a useful answer ...
 
  • #8
g_edgar said:
Woh! a series that diverges for every x ... what a useful answer ...

Yeah, i noticed that after a while...

It MIGHT be, that the formula CAN be used, with extreme caution, since we will mainly use differences between two "values" of the anti-derivative. Thos difference might be convergent, even though both terms are not.

But, then again, a numerical integration scheme might do equally well...
 
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  • #9
apologize for my ignorance but what is that process called and the Sigma looking symbol? I am an Yr 12 student currently doing VCE and studying specialist math and is just stumped on an equation hoping to find an answer in here.

NB: excuse me for fail to type with mathematic symbol
Anti-Differentiate x^x=?

But really i am asking how to anti-differentiate x^x(lnx+1) which comes from the derivative of y=x^x
Because out of curiosity i always hold the belief in math if there is a forward operation there should be a backwards operation so if i can differentiate x^x to get that ugly function to anti-differentitate what operations would i have to undergo.

Thanks for the trouble of reading this passage.
 
  • #10
Please start a new thread next time instead of posting to an already existing thread.

norice4u said:
apologize for my ignorance but what is that process called and the Sigma looking symbol?

The Sigma symbol is the summation symbol. It's just the shorthand for a sum. For example

[tex]\sum_{i=1}^3 i^2= 1^2+2^2+3^2[/tex]

Of course, things like [itex]\sum_{i=1}^{+\infty}[/itex] can not be defined as such since the sum would be infinite. Infinite sums are called series in mathematics and have a very big underlying theory.

NB: excuse me for fail to type with mathematic symbol
Anti-Differentiate x^x=?

This function certainly has an anti-derivative, but it can not be written in terms of elementary functions. Most functions do not have elementary anti-derivatives.

But really i am asking how to anti-differentiate x^x(lnx+1) which comes from the derivative of y=x^x

This integral can be solved by an easy substitution.

Because out of curiosity i always hold the belief in math if there is a forward operation there should be a backwards operation so if i can differentiate x^x to get that ugly function to anti-differentitate what operations would i have to undergo.

The "backwards operation" of differentiation is called integration. But integration is much more harder. Where differentiation has nice algorithms which can be used to differentiate all nice functions, the same is not true for integration. Most integrals are not easy to solve.

Anyway, I am locking this thread. Please make a new thread if you have further questions.
 

1. What is the general formula for calculating the integral of e^x.(lnx)dx?

The general formula for calculating the integral of e^x.(lnx)dx is ∫e^x.lnx dx = e^x(lnx-1)+C, where C is the constant of integration.

2. How do you solve the integral of e^x.(lnx)dx using integration by parts?

To solve the integral of e^x.(lnx)dx using integration by parts, we first identify u=e^x and dv=lnx dx. Then, we use the formula ∫u dv = uv - ∫v du to get the final answer of e^x(lnx-1)-∫e^x/x dx.

3. Is there a specific substitution method for solving the integral of e^x.(lnx)dx?

Yes, we can use the substitution method by letting u=lnx. This will result in the integral becoming ∫e^(e^u) du, which can be solved using the power rule for integration.

4. Can the integral of e^x.(lnx)dx be solved using the u-substitution method?

Yes, the u-substitution method can be used to solve the integral of e^x.(lnx)dx. By letting u=lnx, the integral becomes ∫e^(e^u) du, which can then be solved using the power rule for integration.

5. Is there a specific method for solving the definite integral of e^x.(lnx)dx?

Yes, we can use the definite integrals property to solve the integral of e^x.(lnx)dx. By using the property ∫a^b f(x) dx = F(b)-F(a), where F(x) is the antiderivative of f(x), we can evaluate the definite integral by plugging in the limits of integration into the antiderivative formula.

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