How to Solve the Integral of e^x ln(x) dx?

[ultrazyn]

i need help with this integral

$$\int{e^x}{ln{x}}dx$$

when i let u=ln(x) and v=e^x, i get

$$\{e^x}{lnx}\ - \int{e^x/{x}}dx$$

and then it keeps looping

 

[Tide]

What do you mean by "it keeps looping?"

 

[HallsofIvy]

If by "looping", you mean that you let $dv= \frac{1}{x}$ and $u= e^x$ and everything canceled out leaving you with
[tex]\int e^x lnx dx= \int e^x ln x dx[/itex]<br /> Of course!<br /> <br /> If you do integration by parts twice and the second time "swap" u and dv, that will always happen.<br /> <br /> Perhaps what you mean is that you again let $dv= e^x$ and $u= \frac{1}{x}$ so that you would up with<br /> [tex]e^x lnx dx= e^x lnx- \frac{1}{x}+ \int\frac{e^x}{x^2}dx[/itex]<br /> and continuing just gives you higher and higher powers in the denominator: Okay, that's good. What eventually happens is that, in the limit, you get ln x plus an infinite series. The integral cannot be done in terms of "elementary" functions.[/tex][/tex]

 

[Pythagorean]

If, when solving by parts, you ever end up with your original integral, you can substitute a algebraically (like substitute the letter I for the origingal integral anywhere you see it) then just solve for I and that's your solution.

My professor broke solving by parts into three different types, and he called that one "around the world"

 

[eljose]

If by "looping", you mean that you let $dv= \frac{1}{x}$ and $u= e^x$ and everything canceled out leaving you with
[tex]\int e^x lnx dx= \int e^x ln x dx[/itex]<br /> Of course!<br /> <br /> If you do integration by parts twice and the second time "swap" u and dv, that will always happen.<br /> <br /> Perhaps what you mean is that you again let $dv= e^x$ and $u= \frac{1}{x}$ so that you would up with<br /> [tex]e^x lnx dx= e^x lnx- \frac{1}{x}+ \int\frac{e^x}{x^2}dx[/itex]<br /> and continuing just gives you higher and higher powers in the denominator: Okay, that's good. What eventually happens is that, in the limit, you get ln x plus an infinite series. The integral cannot be done in terms of "elementary" functions.[/tex][/tex]

[tex][tex] <br /> Except for the case $$\int_0^{\infty}dxe^{-ax}lnx=-\frac{\gamma+lna}{a}$$ which is exact.[/tex][/tex]

 

[matt grime]

Your integral is a definite integral, the other is not, moreover, what is gamma?

 

[gnomedt]

Gamma is probably the Euler-Mascheroni Constant (not 2.718...).

http://mathworld.wolfram.com/Euler-MascheroniConstant.html

 

[Orion1]

$$\text{ExpIntegralEi}[x] = - \int_{-x}^\infty \frac{e^{-t}}{t} dt$$

Mathematica solution:
$$\int e^x ln{x} dx = \int_{-x}^\infty \frac{e^{-t}}{t} dt + e^x ln x$$

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$$\gamma$$ - EulerGamma (Euler-Mascheroni Constant)

$$\text{ExpIntegralEi}[z] = \sum_{k=1}^\infty \frac{z^k}{kk!} + \frac{1}{2} \left[ ln z - ln \left( \frac{1}{z} \right) \right] + \gamma$$

$$\int_{-z}^\infty \frac{e^{-t}}{t} dt = \sum_{k=1}^\infty \frac{z^k}{kk!} + \frac{1}{2} \left[ ln z - ln \left( \frac{1}{z} \right) \right] + \gamma$$

Orion1 solution:
$$\int e^x ln{x} dx = \sum_{k=1}^\infty \frac{x^k}{kk!} + \frac{1}{2} \left[ ln x - ln \left( \frac{1}{x} \right) \right] + e^x ln x + \gamma$$

Reference:
http://functions.wolfram.com/GammaBetaErf/ExpIntegralEi/02/
[/Color]
http://mathworld.wolfram.com/Euler-MascheroniConstant.html

 

[ashu_manoo12]

if "lnx" is the anle of any trignometric function. like intergration of "sinln dx "

 

[d_leet]

if "lnx" is the anle of any trignometric function. like intergration of "sinln dx "

What? That made absolutely no sense whatsoever? ln(x) is the natural logarithm of x.