What other methods can be used to integrate e^xlog(x)?

[FedEx]

\int e^xlog(x)

I have solved this sum using integration by parts. The answer which i get is

e^xlogx - logx - \sum^{\infty}_{i=1}\frac{x^i}{(i)(i!)}

But i also used the series expansion of e^x. Is there any other way of doing this sum?

I have almost tried out every single way of doing this sum by parts. So better think about substitution or any other possible method other then by parts.

 

[MathematicalPhysicist]

substitute e^x=u, you get e^xdx=du, dx=du/u, you can fill the blanks by your own.

 

[HallsofIvy]

If you know that \int ln x dx= x ln x - x+ C, which you can do by integration by parts, then it is easy to take u= e^x, dv= ln x and use integration by parts.

 

[FedEx]

substitute e^x=u, you get e^xdx=du, dx=du/u, you can fill the blanks by your own.

But on doing so we will get

log(log u) du to integrate. And i did integrate that but i had to use series expansion of e^x.

 

[FedEx]

If you know that \int ln x dx= x ln x - x+ C, which you can do by integration by parts, then it is easy to take u= e^x, dv= ln x and use integration by parts.

I am sorry but i am not able to get you.

Integration of log x is simple. Here what do you mean by dv? Me being in India have a slight of different version of notations. Do you mean that we take log x as the second function? I have tried that but i get

xe^xlogx to integrate. Which i am unable to do

 

[D H]

substitute e^x=u, you get e^xdx=du, dx=du/u, you can fill the blanks by your own.

That's probably what the OP did, with one error.

If you know that \int ln x dx= x ln x - x+ C, which you can do by integration by parts, then it is easy to take u= e^x, dv= ln x and use integration by parts.

How does this help?

\int e^xlog(x)

I have solved this sum using integration by parts. The answer which i get is

e^xlogx + \frac{1}{x^2} - \sum^{\infty}_{i=1}\frac{x^i}{(i)(i!)}

That looks good except for one term. Could you show your derivation?

This integral does not have a solution in the elementary functions. In other words, the only way to express the integral is to use either series as you have done or to use some special function. If you choose the appropriate values for u and dv, you will get a rather widely used special function in one step.

Note: I'm not saying you should use that particular special function. Your instructor may well want the series representation.

 

[FedEx]

If you choose the appropriate values for u and dv, you will get a rather widely used special function in one step.

I am not able to get this.

Take

x= log\alpha

By parts i get

\alpha log(log\alpha) - \int\frac{d\alpha}{log\alpha}

let log\alpha = \beta

we get

\alpha log(log\alpha) - \int\frac{e^{\beta}d\beta}{\beta}

Then i apply series and integrate

\alpha log(log\alpha) - \frac{1}{{\beta}^2} - \beta - \frac{{\beta}^2}{(2)(2!)} - \frac{{\beta}^3}{(3)(3!)} ...

Now bringing it back in x
<br /> e^xlogx -logx - \sum^{\infty}_{i=1}\frac{x^i}{(i)(i!)}

 

[D H]

Take

x= log\alpha

By parts i get

\alpha log(log\alpha) - \int\frac{d\alpha}{log\alpha}

let log\alpha = \beta

we get

\alpha log(log\alpha) - \int\frac{e^{\beta}d\beta}{\beta}

Wow. That was a hard way to get to this result. Look at what you have done. \beta is just x. You assigned x= \log\alpha and \log\alpha = \beta. By transitivity, \beta=x.

Since you arrived at this result the hard way, here's an easy way: Integrate by parts directly.

Let

\aligned<br /> u&amp;=\log x &amp;\quad dv &amp;=e^x dx \\<br /> du &amp;= \frac {dx}x &amp; \quad v &amp;= e^x \endaligned

Thus

\int e^x\log x dx = e^x\log x - \int \frac{e^x}{x} dx

Then i apply series and integrate

\alpha log(log\alpha) - \frac{1}{{\beta}^2} - \beta - \frac{{\beta}^2}{(2)(2!)} - \frac{{\beta}^3}{(3)(3!)} ...

Too fast! Where does the 1/beta^2 come from?

 

[FedEx]

Ya.I saw it at last while putting back the values in terms of x.

\frac{e^{\beta}}{\beta} = \frac{{1} + \beta + \frac{{\beta}^2}{2!} + \frac{{\beta}^3}{3!} + ...}{\beta}

Then i integrated it.

 

[D H]

What is the integral of 1/beta? (Hint: It is not 1/beta²).

 

[FedEx]

Oh my god. How can this happen. I am so sorry. I am a fool. Its

log{\beta}

I am so sorry. I a ashamed of myself. May be its because i am ill at the moment. However illness has nothing to do with it this is just because of my mere lack of concentration

 

[D H]

Naw. You just made a dumb mistake. We all make dumb mistakes. In general, admit it (eat some crow) and move on. Crow can be a tasty dish at times.

Now that you have the answer in series form, there is a way to express it without a series using the "exponential integral":

\textrm{Ei}(x) \equiv \int_{-\infty}^x \frac {e^t}t dt

 

[FedEx]

Naw. You just made a dumb mistake. We all make dumb mistakes. In general, admit it (eat some crow) and move on. Crow can be a tasty dish at times.

Now that you have the answer in series form, there is a way to express it without a series using the "exponential integral":

\textrm{Ei}(x) \equiv \int_{-\infty}^x \frac {e^t}t dt

Thanks for the consolation.

So you mean that the sigma and the log x term can be replaced by the exponential integral, isn't it?

 

[D H]

Yes, but if you haven't covered special functions yet your instructor probably wants to see the series representation.

 

[FedEx]

Thanks a lot D H. You really helped me with the sum